Physics numericals for class 10 chapter 13 electrostatics
Grade 10 KPK Class 10 Notes Physics numericals for class 10 chapter 13 electrostatics, Conceptual Questions, comprehensive questions, and numerical questions answers.
Conceptual Questions Chapter 13 Physics for Class 10
Table of Contents
Q.1) Normally, objects with large number of electrons are electrically neutral, Why?
Answer: Most objects have the same number of electrons and protons in them, therefore, we have equal and opposite charges and as a result, the net effect is zero and so these objects are electrically neutral. So, objects which have a large number of electrons must also have the same number of protons to be electrically neutral.
Q.2) How does shuffling feet across a carpet cause hair to stand on our body?
Answer: When we shuffle our feet on the carpet, we are rubbing electrons off the carpet and onto our body, these electrons cause the hair on our body to stand.
Q.3) Why neutral objects are always attracted by charged object? Not repelled.
Answer: This is due to electrostatic induction. When an electrically charged object is bought near a neutral object it causes the redistribution of charges in a neutral object. Let suppose the charged object is negative than positive charges in the neutral object will move towards the region closer to negatively charge object and the negative charges will move towards the further region. The positive charges of the neutral object will attract the negatively charged object, as a result, the natural object will get attracted towards the charged object. Similarly, if a positively charged object is brought near a neutral object then the negative charges in the neutral object will move closer to the charged object and positive will move further. Hence the charged object will attract the neutral object.
Q.4) Why the pieces of Paper initially attracted by charged comb fly away when they touch it?
Answer: The charged comb attracts paper pieces of paper due to the electrostatic induction. When a comb is brought near the pieces of paper, a charge redistribution occurs in the paper pieces. Charges in the pieces of paper with the same sign are pushed away while charges with the opposite sign are pulled closer. This change in distance means the attractive force due to the closer opposite sign charges is greater than the repulsion due to same sign charges which have been pushed further away. This creates a net attractive force which pulls the papers towards the comb. When the comb touches the pieces of paper, it gets neutralized and there is no more attraction between them, as a result, the pieces fly away.
Q.5) Is it necessary for a charged body actually to touch the ball of the electroscope for the leaves to diverge? Defend your answer.
Answer: It is not necessary to touch the ball of electroscope to diverge the leaves of the electroscope. When a charged object is brought near the ball of the electroscope, due to the electrostatic induction charges in the ball, metal rod, and leaves are redistributed. The ball will acquire the charge opposite to that of the charged object and the rod and leaves will acquire the same charge of the charged object. Now the leaves and rod have charges due to which they will deflect each other. Hence the leaves of the electroscope can be diverged without touching the ball.
Q.6) How electrostatic painting is better than conventional spray painting?
Answer: Electrostatic painting is a method in which an electrostatically charged paint is applied. A body is charged and then the paint is given the opposite charge by charging the nozzle of the sprayer. Due to the mutual repulsion charge particles are pushed out of the nozzle and form a mist which is attracted towards the oppositely charged body. This method reduce the paint usage and uneven coating that result from using regular conventional spray painter, both for powder and liquids paint.
Q.7) Why are lightning rods normally at a higher elevation than the buildings they protect?
Answer: The lightning rods are normally installed at a higher elevation than the building to protect the building against direct lighting strikes. By placing the lightning rod at higher elevation, the lightning will struck the rod first instead of the building and the current will discharge to the earth. This way the building will be protected from the direct strike of lightning.
Q.8) What would happen if two insulating plates were used instead of conducting plates to construct a capacitor?
Answer: In ideal capacitor, conducting plates are used to store the charges, the charges get spread across the plates, the plate connect to negative terminal of battery becomes negatively charged and the plate connect to positive terminal gets positively charged. If the plates are replaced by insulator, then there is no free electron to move and cause charge separation. Hence we would store no charge. So it will not act like a capacitor.
Q.9) The sum of the charges on both plates of a Capacitor is zero. What does a capacitor store?
Answer: A capacitor has two plate having equal and opposite charge, as a result the total sum of the charges on the capacitor is zero. Due to the opposite charges, a potential difference is generated between the plates and there exist a electric field between them. The electric field stores the electric energy. Hence we can say capacitor stores energy.
Q.10) If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.
Answer: To store a large amount of charge in capacitor bank the capacitors should be connected in parallel. Explanation: The capacitance of a capacitor defines how much charge a capacitor can store. In the case of series, the equivalent capacitance of the capacitor bank is given by 1/Ce = 1/C1 + 1/C2 + 1/C3 + …… + 1/Cn While for the case of parallel combination the equivalent capacitance of the capacitor bank is Ce = C1 + C2 + C3 + ….. + Cn We can see from relations that the equivalent capacitance in parallel is always greater than that of the series. Hence due to the larger capacitance, the capacitor bank formed by the connecting the capacitors in parallel will store a large amount of charge.
comprehensive questions physics for class 10 chapter 13 electrostatics
Q.1) What is electric charge? How objects can be electrified? Describe with the help of experiments.
Answer: Electric charge: Electric charge is a basic property of a material body due to which it attracts or repels another object. The SI unit of charge is coulomb (C). Electrification:
In the above figure, we can see that similar charges repel and different charges attract. Electric charge is not created in the process of charging objects, charges are only transferred between the objects. Objects can be charged by removal or addition of charges (specifically electrons) called electrification. Since electrons can be transferred easily therefore if an object has a; 1. Positive (+) charge – means it has less electrons than normal. 2. Negative (-) charge – means it has more electrons than normal.
Q.2) What is electrostatic induction? Explain.
Answer: Answer: Electrostatic Induction: “The modification of distribution of charge on a material under the influence of an electric charge on a nearby object is called the electrostatic induction” In this phenomena all induced charge remain static and that is the reason that it is known as electrostatic induction. Induction of Charge in a Conductor: Consider a metallic sphere fixed on an insulated wooden stand as shown in figure. Let we bring a negatively charged rod close to one end of metallic sphere say end A. Then due to this negatively charged rod, all positive charges of sphere will accumulate to its end A due to electrostatic force of attraction between oppositely charged particles. However all negative charge will accumulate at the other end i.e. at end B. Total amount of charge at both ends will be equal. But when we remove the charged rod then all accumulated charges from the ends will disappear. In this way we can induce a charge in a conductor.
Induction of Charge in an Insulator: Charge can be induced on an insulator by running it to another material. Consider the rubbing of a comb in hairs. As a result of this rubbing, a negative charge will produce on the surface of comb. This induction of charge can be visualized by bringing this comb close to some paper pieces. On getting closer, comb will repel the negatively charged particles of paper pieces away and the positively charged particles will accumulate near comb end. Therefore, the comb will attract these paper pieces. Here both, the comb and paper pieces are insulators but they get charged due to effect of electrostatic induction.
Q.3) What is the function of electroscope? How can we use electroscope to find the presence and nature of charge on a body?
Answer: Electroscope: An electroscope is an instrument used for detecting and testing the nature of charge on a body.It consists of a glass jar in which a metal rod is fitted. This metal rod which has a metallic disc at its upper end and a gold, silver, tin or other metal leaf at the other end. Both, the rod and the jar are totally detached from each other. We usually use electroscope in order to detect the presence of charge on a body and it also gives us information about the nature of the charge. Detection of Charge: If we want to detect the presence of charge on a body then we will bring this body near the disc of electroscope which is neutral at start as shown in figure. Now if there is any kind of charge on the body, which we bring close to the disc then the gold leaf of the electroscope will diverge to some extent. This divergence of leaves is due to effect of electrostatic induction due to charged body. If there was no charge on the body then there will be no divergence of eaves and eaves will remain at their position.
Q.5) What is meant by electric field and electric field intensity? How the field lines represent the electric field for isolated positive and negative point charges?
Answer: Electric Field: ” The region around a charge in which if a test charge is brought will feel an electric force “. It is a vector filed, therefore, the test charge will feel a directed force at each point in that region. Electric Lines of Force: Electric field lines represent the direction and magnitude of the electric field. 1. For a positive charge, the electric field lines will be directed outward as shown in figure (a). 2. For negative charge, direction of electric field lines will be inward as shown in figure (b).
3. If we place two equal charges of the same charge say two positive charges of equal magnitude then the field lines will have the pattern shown in figure (c). 4. If we place two oppositely charges particles close to each other then their electric filed lines will originate from positive charge and will enter to terminate at a negative charge as shown in figure (d).
5. If two oppositely charges plates are placed parallel to each other at some distance then their electric field will be uniform at the central region of plates. Shown in figure (e).
Electric Field Intensity: The strength of the electric field at any point is known as electric field intensity. In order to find the electric field intensity at a certain point, we place a test charge at that point. If ‘F’ is the force experienced by that test charge then electric field intensity will be given as,
Therefore we can also define electric field intensity as, “The amount of force acting on a unit positive charge at a point in an electric field is known as electric field intensity at that point” SI unit of electric field intensity is NC-1. Electric Lines of Force: Electric field lines represent the direction and magnitude of the electric field. 1. For a positive charge, the electric field lines will be directed outward as shown in figure (a). 2. For negative charge, the direction of electric field lines will be inward as shown in figure (b).
Q.6) What is electric potential? in what units we measure electric potential?
Answer: Electric Potential: The electric potential energy ‘U’ per unit charge ‘q’ in an electric field is called electric potential V. V = U/q = W/q Unit: The electric potential unit is joules per coulomb (JC-1), or volt (V). ” The potential at a point is one volt when it requires one joule of work to move a positive charge of one coulomb from a point of ZERO potential to that point “. 1 volt = 1 joule / 1 coulomb The concept of electric potential is closely related to the electric field. Electric field is the force per unit charge, whereas the electric potential is the energy per unit charge. However, the electric potential is a scalar quantity. Since it is easier to solve a problem with scalars, therefore, it is simpler to solve problems with electric potential rather than the electric field.
Q.7) Give some practical applications in which electric field is useful.
Answer: Electrostatic phenomena have wide application in daily life. A. Electrostatic precipitation and Dust extraction: Electrostatic phenomena can be used to separate dust from smoke particles. To reduce air pollution, modern day coal burning power stations extract dust from the smoke in chimneys before releasing it to the environment by a process called electrostatic precipitation. For this purpose, chimneys have a highly positively charged grid (usually a wire gauze) and negatively charged plates. When smoke rising from chimney containing smoke and dust particles pass through the positively charged grid they acquire a positive charge. These charged particles are attracted by the negatively charged plate and are deposited on them. Thus, the smoke coming out of the chimney is free from dust and other particles. B. Electro painting: Electrostatic spray painting is a method in which electrostatically charged paint is applied. This method reduces paint usage and uneven coating that results from using a regular spray painter, both for powder and liquid paint. One type of system applies a negative electric charge to the paint while it is in the container. Other systems apply the charge in the barrel of the spray painter gun. The paint is then pushed through the gun, rubbing against the side, and gaining a static electric charge as it moves. Since the paint particles all have the same charge, they repel each other. This helps to distribute the paint particles evenly and get uniform coverage.
Q.8) How lightning occurs? How can we safeguard ourselves from Lightning hazard?
Answer: Lightning: Apart from useful applications of electrostatics – it can sometimes be dangerous. For example, lightning is the result of large scale charge separation occurring within a thundercloud, Lightning involves the dielectric breakdown of air. Charge separation occurs within thundercloud; the top of the cloud becomes positive and the lower part becomes negative. The negative charge at the bottom of the thundercloud induces a positive charge on the Earth just underneath the cloud. When the electric field between the cloud and the earth becomes large enough, the air undergoes dielectric breakdown, meaning it momentarily becomes a good conductor of electricity allowing the negative charge to jump from the cloud to the earth. A lightning channel is completed and electrons rush to the ground making the channel glow in the process. A total of about 20 °C to 25 °C of electronic charge is transferred from the thundercloud to the surface. Safety Precautions: We should stay indoors or in an automobile if possible. When caught in the open, we should keep low, stay away from any tall tree, if lightning strikes the tree, charge traveling down the tree and then along the surface will put us in danger. If trapped in such a situation we should try to go in a nearby ditch or low spot keeping our head low and feet as close together as possible.
Q.9) What is capacitor? Define capacitance and its units.
Answer: Capacitor: ” A device used for storing charge, it consists of two conductors separated from (without touching) each other, carrying charges of equal magnitude but an opposite sign “. The figure below shows the basic elements of any capacitor- two isolated conductors (having charge +Q and -Q ) of any shape. No matter what their geometry, flat or not, we call these conductors plates. Each capacitor plate carries a charge of the same magnitude, one positive and the other negative. Due to the charges, the electric potential of the positive plate exceeds that of the negative plate by an amount V. Capacitance: If we provide some charge Q to one plate of the capacitor then there will be a potential difference V between the plates. If we increase the charge on plate then the potential difference will also increase Therefore we can say that charge and potential difference are directly proportional to each other in case of a capacitor. Hence, we can write
Here ‘C’ is the constant of proportionality, also known as the capacitance of capacitor. The capacitance of a capacitor is given as,
Here ‘C’ is the constant of proportionality, also known as the capacitance of capacitor. The capacitance of a capacitor is given as,
The value of capacitance ‘C’ depends on the area of plates, the distance between these plates and the nature of medium between them. Unit: SI unit of capacitance is called Farad and is denoted by ‘F’. “If a capacitor stores a charge of 1 coulomb and a potential difference between the plates is 1 volt then the capacitance of the capacitor will be one Farad”
Q.10) What is series combination of capacitors? How we can determine equivalen capacitors for different capacitors connected in series?
Answer: Series Combination of Capacitors: From the figure below, we can understand the series combination of capacitors. All plates of three capacitors are connected to each other plate to plate. Left plate of first capacitor is connected to point A and the right plate of third capacitor is connected to point B. Let we apply a voltage V across points A and B such that positive terminal of the battery is connected to point A and negative terminal to point B. This combination is known as a series combination of capacitors.
When a battery is connected to a series combination of capacitors, the same current flows through each capacitor which means a charge of +Q is placed on the left plate of each capacitor and an equal charge of -Q on the right plate of each capacitor. As a result, each capacitor gets an equal amount of charge Q on each of its plates. Q1 = Q2 = Q3 = Q ……….. …………. ……….. (1) When the three capacitors in the circuit are charged, the sum of the potential drops across all three must equal the potential difference supplied by the battery. V = V1 + V2 + V3 ………. ………… ………… (2) Since the capacitance of the capacitor is
Therefore, each voltage can be written as
Where Ce is the equivalent capacitance of a single capacitor that has the same effect on the circuit as the series combination when it is connected to the battery. Hence equation (2) can be written as,
From equation (1) in equation (3) can be written as
Generally for ‘n’ number of capacitors connected in series
From the above equations, it is clear that in series combination, the equivalent capacitance is always smaller than individual capacitance in combination.
“physics numericals for class 10 chapter 13 electrostatics”
Q.1) Determine the magnitude of the electric force on the electron of a hydrogen atom exerted by the single proton that is the atom’s nucleus. Assume the average distance between the electron and the proton is r = 5.3 x 10-11 m and charge on electrons and proton is 1.6 x 10-19 C.
Answer: Given data: Distance = r = 5.3 × 10-11 m Charge on electron = charge on proton = q = 1.6 × 10-19 C Columb constant = k = 9× 109 NC-2m2 To find: Force = F =? Calculation: As we know
Q.2) A 5 μC point charge is placed 20cm from a 10 μC point charge. (a) Calculate the force experienced by the 5 μC charges. (b) What is the force on the 10 μC charges? (c) What is the field strength 20 cm from the 10 μC point charge?
Answer: Given data: First Charge = q1= 5 μC = 5 × 10-6 C second charge = q2 = 10 μC = 10 × 10-6 C Distance between q1 and q2 = r = 20 cm = 0.2 m Columb constant = k = 9 × 109 NC-2m2 To find: a. Force on the first charge = F21 =? b. Force on second charge = F12 =? c. Field strength from the second charge = E =? Calculation: a.
b. Since, F21 = F12 so, F12 = 11.25 N c. The field strength is given by, E = F12/q E = 11.25/10 × 10-6 E = 1125000 N/C
Q.3) In a certain region of space, a uniform electric field has a magnitude of 4.60 x 104 N/C and points in the positive x-direction. Find the magnitude and direction of the force this field exerts on a charge of (a) +2.80 μC (b – 9.30 μC.
Answer: Given data: Electric field = E = 4.60 × 104 N/C First Charge = q1 = 2.80 μC =2.80 × 10-6C second charge = q2 = -9.30 μC = -9.30 × 10-6 C To find: Magnitude and direction of force on first charge = F1 = ? Magnitude and direction of force on second charge = F2 =? Calculation: a. We know that F = qE or F1 = q1E F1 = (2.80× 10-6)(4.60 × 104) F1 = 12 . 88 N Direction : Along the postitive x-direction. b. F2 = q2E F2 = (-9.30 × 10-6)(4.60 × 104) F2 = -0.427 N Direction : Along the negitive x-direction.
Q.4) The potential difference between two points is 110 V. When an unknown charge is moved between these two points, the work done is 550 J. What is the amount of charge?
Answer: Given data: Potential difference = V = 110 V Work done = W = 550 J To find: Charge = q =? Calculation: The work done a charge due a potential difference is, W = qV or q = W/V q = 550/110 q = 5 C
Q.5) The capacitance of a capacitor is 3200 pF. If the potential difference between its plates is 220 V. What is the charge on each of its plates?
Answer: Given data: Capacitance = C = 3200 pF = 3200 × 10-12 F Potential difference = V = 220 V To find: Charge = Q = ? Calculation: The charge stored on the plates of a capacitor is Q = CV Q = (3200 × 10-12)(220) Q = 0.000000704 C Q = 7.04 × 10-7 C
Q.6) Three capacitors of capacitance 1 μF, 2 μF and 3 μF are connected in Series to a 110 V battery. Calculate the equivalent capacitance and ‘Voltage across each capacitor.
Answer: Given data: Capacitance = C1 = 1 μF Capacitance = C2 = 2 μF Capacitance = C3 = 3 μF Voltage = V = 110 V To find: Equivalent capacitance = Ceq = ? Voltage across C1 = V1 = ? Voltage across C2 = V2 = ? Voltage across C3 = V3 = ? Calculation: For series combination of capacitors the equivalent capacitance is,
Now for charge stored on the capacitor. Q = Ceq V Q = 0.54 (110) Q = 59.4 C Now Voltage across C1 : V1 = Q1/C1 Since the charge remains the same so Q1 = Q2 = Q3 = Q V1 = Q / C1 V1 = 59.4 / 1 V1 = 59.4 V Voltage across C2 : V2 = Q / C2 V2 = 59.4 / 2 V2 = 29.7 V Voltage across C3 : V3 = Q / C3 V3 = 59.4 / 3 V3 = 19.8 V
Q.7) Two capacitors of capacitance 2 pF and 3 pF are connected in parallel to 9V battery. Calculate the equivalent capacitance and the charge on each capacitor.
Answer: Given data: Capacitance = C1 = 2 pF = 2 × 10-12 F Capacitance = C2 = 3 pF = 3 × 10-12 F Voltage = V = 9V To find: Equivalent capacitance = Ceq = ? Charge on C1 = Q1 = ? Charge on C2 = Q2 = ? Calculation: For parallel combination of capacitor we know Ceq = C1 + C2 Ceq = 2 × 10-12 + 3 × 10-12 Ceq = 5 × 10-12 Ω Now Charge on C1: Q1 = C1V1 Since in parallel combination, V = V1 = V2 So, Q1 = C1V Q1 = (2 × 10-12)(9) Q1 = 1.8 × 10-11 C Charge on C2: Q2 = C2V Q2 = (3 × 10-12)(9) Q2 = 2.7 × 10-11 C