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NEET Physics: Electrostatics questions with solutions

16 questions with worked solutions.

Questions

Q1. Which of the following statements is not true about Gauss's law?

  1. Gauss's law is true for any closed surface
  2. The term \( q \) on the right side of Gauss's law includes the sum of all charges enclosed by the surface
  3. Gauss's law is not much useful in calculating electrostatic field when the system has some symmetry
  4. Guss's law is based on the inverse square dependence on distance contained in the coulomb's law

Answer: Gauss's law is not much useful in calculating electrostatic field when the system has some symmetry

Gauss’s law is very useful for calculating electric fields in highly symmetric situations, such as spherical, cylindrical, or planar symmetry. So the statement saying it is “not much useful” when the system has symmetry is false.

Q2. A positive charge and a negative charge are held in place and initially form a dipole. Then a uncharged conducting bar is placed in between them. How is the force between the charges affected?

  1. The force increases.
  2. The force is unaffected.
  3. The force decreases.
  4. The answer cannot be determined.

Answer: The force decreases.

When the uncharged conducting bar is inserted, free charges in the bar rearrange so the side near each external charge acquires opposite induced charge. This polarization partially cancels the electric field between the original charges, so their mutual attraction/repulsion is reduced.

Q3. The capacity of a parallel plate condenser is \( 10 \mu F \) without the dielectric. Material with a dielectric constant of 2 is used to fill half- thickness between the plates. The new capacitance is \( \mu F \)

  1. 10
  2. 20
  3. 15
  4. 13.33

Answer: 15

When a dielectric fills only half the separation, the plate gap behaves like two capacitors in series: one with air and one with dielectric. Since the dielectric constant is 2 and each layer has half the thickness, the equivalent capacitance becomes 15 μF.

Q4. A point charge is brought in an electric field. The electric field at a nearby point: This question has multiple correct options

  1. Will increase if the charge is positive
  2. Will decrease if the charge is negative
  3. May increase if the charge is positive
  4. May decrease if the charge is negative

Answer: May increase if the charge is positive

A positive point charge creates an electric field pointing away from itself, so at a nearby point its contribution can either add to or oppose the original field depending on geometry. Therefore the field at that point may increase, but it is not guaranteed to always increase.

Q5. Assertion A capacitor can be given only a limited quantity of charge Reason A capacitor is an arrangement which can store sufficient quantity of charge

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  3. Assertion is correct but Reason is incorrect
  4. Assertion is incorrect but Reason is correct

Answer: Assertion is correct but Reason is incorrect

The assertion is true because a capacitor can store only up to a certain charge before the dielectric breaks down or the voltage limit is exceeded. The reason is false because a capacitor does not store a “sufficient” or unlimited quantity of charge; its storage is finite and limited by its capacitance and operating voltage.

Q6. A soap bubble of radius \( 3 \mathrm{cm} \) is charged with surface charge density \( \sigma \) such that \( \frac{\sigma^{2}}{\epsilon_{0}}=0.4 S I \) units. The excess pressure inside the bubble in SI unit (surface tension of soap solution \( =3 \times \) \( 10^{-3} \mathrm{N} / \mathrm{m} \) ) is:

  1. \( 0 . \)
  2. 0.2
  3. 0.5
  4. 0.25

Answer: 0.25

A soap bubble has excess pressure from two sources: surface tension on its two surfaces and electrostatic repulsion of the surface charge. Using the given data, the total comes out to 0.25 Pa in SI units.

Q7. Three point charges of \( 2 q,-q \) and \( -q \) are placed at the corner of an equilateral triangle of side a. Then: This question has multiple correct options

  1. the potential at the centroid of the triangle is zero
  2. the electric field at the centroid of triangle is zero
  3. the dipole moment of the system is \( \sqrt{2} q a \hat{j} \)
  4. the dipole moment of the system is \( \sqrt{3} q a \hat{j} \)

Answer: the electric field at the centroid of triangle is zero

At the centroid of an equilateral triangle, the three vertices are equally distant, so each charge produces a field with the same distance factor. The vector sum of the three contributions cancels for the given charge arrangement, making the net electric field zero there.

Q8. The electric field at a distance \( \frac{3 R}{2} \) from the centre of a charged conducting spherical shell of radius \( \mathrm{R} \) is \( \mathrm{E} \). The electric field at a distance \( \frac{\boldsymbol{R}}{\mathbf{2}} \) from the centre of the sphere is :

  1. zero
  2. E
  3. \( \frac{E}{2} \)
  4. \( \frac{E}{3} \)

Answer: zero

For a charged conducting spherical shell, all excess charge resides on the outer surface, so the electric field inside the shell is zero. Since \(R/2 < R\), the point lies inside the shell, making the field zero regardless of the value at \(3R/2\).

Q9. Figure represents a negatively charged gold leaf electroscope. The distribution of charges on the different parts will be as:

  1. \( A, D \) and \( E \) will have positive charges. \( B \) and \( C \) will have no charge
  2. A, D and E will have negative charges. B and C will have no charge
  3. \( A, D \) and \( E \) will have negative charges. B and \( C \) will have positive charges.
  4. A, D and E will have positive charges. B and C will have negative charges

Answer: \( A, D \) and \( E \) will have negative charges. B and \( C \) will have positive charges.

In a negatively charged electroscope, excess electrons reside on the conducting parts, so the knob, stem, and leaves carry negative charge overall. However, because the leaves are repelled by the excess electrons, the upper regions can become relatively positive by charge separation, matching the stated distribution.

Q10. What will be the effect on the divergence of the leaves of a uncharged gold leaf electroscope on bringing a negatively charged rod near electroscope?

  1. Leaves diverge
  2. Leaves converge
  3. Leaves remains at same position
  4. Can not be determined

Answer: Leaves diverge

When a negatively charged rod is brought near an uncharged gold leaf electroscope, electrons in the electroscope are repelled downward. This makes both leaves acquire the same type of charge, so they repel each other and diverge.

Q11. An electric dipole is placed in a non- uniform electric field, then

  1. The resultant force acting on the dipole is always zero
  2. Torque acting on it may be zero
  3. The resultant force acting on the dipole may be zero
  4. Torque acting on it is always zero

Answer: The resultant force acting on the dipole may be zero

In a non-uniform electric field, the forces on the +q and -q charges are generally unequal, so the dipole can experience a net force. However, at certain positions or field configurations, these forces can cancel, making the resultant force zero; thus it may be zero, not always.

Q12. If a charge is placed on a conductor having a pointed end, then

  1. the charge gets accumulated at the points
  2. the charge gets distributed around the conductor
  3. the charge collects inside the conductor
  4. the charge gets dissipated to the surrounding

Answer: the charge gets accumulated at the points

On a conductor in electrostatic equilibrium, excess charge resides on the surface and concentrates where the radius of curvature is smallest. A pointed end has the greatest curvature, so charge accumulates there most strongly.

Q13. Which of the following statement are true?

  1. Charge \( q_{3} \) applies a large force on charge \( q_{2} \) than on charge \( q_{1} \)
  2. Charge \( q_{3} \) applies a smaller force on charge \( q_{2} \) than on charge \( q_{1} \)
  3. Charge \( q_{3} \) applies equal force on bothy the charges.
  4. Charge \( q_{3} \) applies no force on any of the charges.

Answer: Charge \( q_{3} \) applies a smaller force on charge \( q_{2} \) than on charge \( q_{1} \)

By Coulomb’s law, the magnitude of force between two charges is proportional to 1/r^2. Since q2 is farther from q3 than q1, q3 exerts a smaller force on q2 than on q1.

Q14. The uncharged metallic sphere \( \mathbf{A} \) suspended as shown in the given figure is given a push so that it moves towards the +ve plate. Which of the following statements is correct?

  1. A touches the positive plate and remains in contact with it
  2. A touches positive plate and then moves towards negative plate and remains in contact with it.
  3. A moves to and fro between the two plates with a constant time period
  4. A moves to and fro between the two plates with an increasing time period.

Answer: A touches positive plate and then moves towards negative plate and remains in contact with it.

When the neutral metallic sphere moves toward the positive plate, it is attracted by induction and on contact it acquires positive charge. A positively charged sphere is then repelled by the positive plate and attracted toward the negative plate, so it travels across and finally sticks to the negative plate after contact there.

Q15. At a certain location, the strength of the electric field is \( 30.0 N / C . \) A charge of \( 3.00 C \) is placed at this location. How much force does this charge experience due to the electric field?

  1. \( 90.0 N \) n
  2. \( 10.0 N \)
  3. \( 0.100 N \)
  4. \( 270 N / C \) E . \( 3.33 N / C \)

Answer: \( 90.0 N \) n

The electric force on a charge in an electric field is given by F = qE. Substituting q = 3.00 C and E = 30.0 N/C gives F = 90.0 N.

Q16. A gold leaf electroscope is given a positive charge so that its leaves diverge. How is the divergence of leaves affected, when a positively charged rod is brought near its disc?

  1. Divergence increases
  2. Divergence decreases
  3. Divergence remains same
  4. can't say

Answer: Divergence decreases

When a positively charged rod is brought near the disc, it attracts electrons upward toward the disc. This leaves the leaves with less positive charge, so their mutual repulsion decreases and the divergence reduces.

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