Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero. Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces. In the light of the above statements, choose the most appropriate answer from the options given below :

(A) is correct but (R) is not correct

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

(A) is not correct but (R) is correct

Electrostatics PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q74 with Solution

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Correct answer: A

Assertion (A): Work done on an equipotential surface

An equipotential surface is a surface on which electric potential is the same at every point.

If a charge $q$ moves from one point to another on the same equipotential surface, then

\Delta V = 0

So, the work done by the electric field is

W = q(V_i - V_f) = -q\Delta V = 0

Hence, Assertion (A) is correct.

Reason (R): Electric field lines and equipotential surfaces

Electric field lines are always perpendicular to equipotential surfaces.

This is because if the electric field had any component along the equipotential surface, then moving a charge along the surface would require non-zero work, which would change the potential. That is not possible on an equipotential surface.

Mathematically,

dW = q\vec E \cdot d\vec l

For motion along an equipotential surface, $dV=0$ , so $dW=0$ . Therefore,

\vec E \cdot d\vec l = 0

which means $\vec E$ is perpendicular to the displacement along the surface. Thus, electric field lines are perpendicular to equipotential surfaces.

Hence, Reason (R) is also correct.

Does (R) explain (A)?

Yes. Since the electric field is perpendicular to the equipotential surface, any displacement $d\vec l$ along that surface is perpendicular to $\vec E$ . Therefore,

dW = q\vec E \cdot d\vec l = qEdl\cos 90^\circ = 0

So the reason directly explains why work done is zero.

Correct option

Therefore, both (A) and (R) are correct and (R) is the correct explanation of (A).

So, the correct answer is:

\boxed{\text{A}}

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Correct answer: A

Assertion (A): Work done on an equipotential surface

An equipotential surface is a surface on which electric potential is the same at every point.

If a charge $q$ moves from one point to another on the same equipotential surface, then

\Delta V = 0

So, the work done by the electric field is

W = q(V_i - V_f) = -q\Delta V = 0

Hence, Assertion (A) is correct.

Reason (R): Electric field lines and equipotential surfaces

Electric field lines are always perpendicular to equipotential surfaces.

Mathematically,

dW = q\vec E \cdot d\vec l

For motion along an equipotential surface, $dV=0$ , so $dW=0$ . Therefore,

\vec E \cdot d\vec l = 0

which means $\vec E$ is perpendicular to the displacement along the surface. Thus, electric field lines are perpendicular to equipotential surfaces.

Hence, Reason (R) is also correct.

Does (R) explain (A)?

Yes. Since the electric field is perpendicular to the equipotential surface, any displacement $d\vec l$ along that surface is perpendicular to $\vec E$ . Therefore,

dW = q\vec E \cdot d\vec l = qEdl\cos 90^\circ = 0

So the reason directly explains why work done is zero.

Correct option

Therefore, both (A) and (R) are correct and (R) is the correct explanation of (A).

So, the correct answer is:

\boxed{\text{A}}

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