101. A row in the table gives the correct connection between field situation and equipotential surface.
| Row | Field or source situation | Equipotential surface |
|---|
| P | Single point charge | Concentric spheres centred at the charge |
| Q | Uniform electric field | Spheres centred at every point of the field |
| R | Single point charge | Planes parallel to the radial field everywhere |
| S | Uniform electric field | Closed spheres around the negative side only |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row P
ⓓ. Row S
Correct Answer: Row P
Explanation: For a single point charge, potential depends only on distance \(r\) from the charge. Constant \(r\) in three dimensions gives concentric spherical equipotential surfaces, so row P is correct. In a uniform electric field, equipotential surfaces are planes perpendicular to the field direction, not spheres. A point charge does not have equipotential planes parallel to the radial field. The sign of the charge changes the sign of potential but not the spherical nature of the equipotential surfaces. Correct surface shape is decided by how \(V\) varies in space.
102. A positive charge is moved from point \(P\) to point \(Q\), both lying on the same equipotential surface. The electric field may be non-zero along the region, yet the electrostatic work is zero because:
ⓐ. The endpoints have no potential difference
ⓑ. Work is always zero for every motion of a positive charge
ⓒ. Electric field cannot exist near an equipotential surface
ⓓ. The charge becomes neutral during the motion
Correct Answer: The endpoints have no potential difference
Explanation: If \(P\) and \(Q\) lie on the same equipotential surface, then \(V_Q-V_P=0\). The electrostatic work is \(W_{\text{field}}=q(V_P-V_Q)\), so it becomes zero for movement between these points. This does not require the electric field itself to vanish in the surrounding region. For example, near a point charge, an equipotential sphere exists even though the radial electric field is non-zero. The charge does not become neutral; the zero work comes from zero potential difference. Work depends on the component of force along displacement, not merely on whether a field exists nearby.
103. Read the graph description below.
A two-dimensional drawing shows several curves representing equal potential values around a point charge. Curves farther from the charge are labelled with smaller magnitudes of \(V\), taking \(V=0\) at infinity.
What does each curve in this drawing represent?
ⓐ. A line on which force is zero at every point
ⓑ. An equipotential line in a plane drawing
ⓒ. A path along which potential changes most rapidly
ⓓ. A field line showing the direction of \(\vec{E}\)
Correct Answer: An equipotential line in a plane drawing
Explanation: In two-dimensional drawings, equipotential surfaces are often shown as equipotential lines or curves. Each curve represents points having the same potential value. Around a point charge, the full three-dimensional equipotential surfaces are spheres, while a plane drawing shows them as circles or curves. Field lines are different because they show the direction of \(\vec{E}\), not constant \(V\). The path of most rapid potential change is along the electric field direction, not along an equipotential line. A line of constant potential can exist where the electric field is non-zero.
104. Two points lie on the same equipotential surface of potential \(40\,\text{V}\). A charge \(-3\,\text{C}\) is moved slowly from one point to the other. The change in potential energy is:
ⓐ. \(+120\,\text{J}\)
ⓑ. \(0\,\text{J}\)
ⓒ. \(-120\,\text{J}\)
ⓓ. \(-40\,\text{J}\)
Correct Answer: \(0\,\text{J}\)
Explanation: \( \textbf{Given condition:} \) Both points are on the same equipotential surface of \(40\,\text{V}\).
\( \textbf{Initial and final potentials:} \)
\[
V_1=40\,\text{V},\qquad V_2=40\,\text{V}
\]
\( \textbf{Potential difference:} \)
\[
\Delta V=V_2-V_1=0
\]
\( \textbf{Potential-energy change:} \)
\[
\Delta U=q\Delta V
\]
\( \textbf{Substitution:} \)
\[
\Delta U=(-3\,\text{C})(0\,\text{V})
\]
\( \textbf{Result:} \)
\[
\Delta U=0\,\text{J}
\]
\( \textbf{Final answer:} \) The change in potential energy is \(0\,\text{J}\).
Even for a negative charge, zero potential difference gives zero change in potential energy.
105. Assertion: No work is done by the electrostatic field when a charge moves along an equipotential surface.
Reason: Along an equipotential surface, the potential difference between any two points is zero.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, but Reason does not explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because electrostatic work between two points is linked with potential difference. For motion from point \(1\) to point \(2\), field work is \(W_{\text{field}}=q(V_1-V_2)\). On an equipotential surface, \(V_1=V_2\). Therefore, the field work is zero. The Reason states exactly the condition that makes this work vanish. The result applies to positive and negative charges because the factor multiplying \(q\) is zero.
106. Equipotential surfaces of different potential values cannot intersect because:
ⓐ. Electric field must be zero wherever two surfaces meet
ⓑ. Potential becomes a vector at the crossing point
ⓒ. Work is always positive at an intersection
ⓓ. One point cannot have two potential values
Correct Answer: One point cannot have two potential values
Explanation: If two equipotential surfaces with different potential values intersected, the point of intersection would belong to both surfaces. That would assign two different values of potential to the same point. A unique electrostatic potential cannot have two values at one point for the same reference choice. This is why equipotential surfaces of different \(V\) do not intersect. The reason is not that electric field must vanish there. The problem is the contradiction in the scalar value of \(V\) at a single point.
107. The best comparison between an electric field line and an equipotential line in a two-dimensional drawing is:
ⓐ. Both always show the same curve with the same meaning
ⓑ. A field line shows direction of \(\vec{E}\), while an equipotential line joins points of equal \(V\)
ⓒ. Both can intersect freely because they are only drawings
ⓓ. A field line joins equal potential points, while an equipotential line shows force direction
Correct Answer: A field line shows direction of \(\vec{E}\), while an equipotential line joins points of equal \(V\)
Explanation: An electric field line is drawn so that its tangent gives the direction of the electric field \(\vec{E}\). An equipotential line is a two-dimensional representation of points having the same potential \(V\). These are different types of information about the same electrostatic region. Field lines are about direction of force on a positive test charge, while equipotential lines are about constant work-per-charge value. They should not be treated as two names for the same curve. In electrostatic diagrams, their relationship becomes especially useful because field direction is perpendicular to equipotentials.
108. A region contains three marked surfaces \(S_1\), \(S_2\), and \(S_3\). A charge moved along \(S_1\) needs no electrostatic work, along \(S_2\) also needs no electrostatic work, and along \(S_3\) also needs no electrostatic work. The most reasonable classification is:
ⓐ. All three are equipotential surfaces
ⓑ. All three must have the same shape and same potential value
ⓒ. All three must be electric field lines
ⓓ. All three must be surfaces where electric field is zero
Correct Answer: All three are equipotential surfaces
Explanation: If no electrostatic work is done while moving a charge along a surface, the potential difference between any two points on that surface is zero. That is the defining feature of an equipotential surface. Different equipotential surfaces may have different potential values, such as \(20\,\text{V}\), \(10\,\text{V}\), and \(0\,\text{V}\). They also need not always have the same shape; the shape depends on the source arrangement or field pattern. Zero work along the surface does not require zero electric field everywhere. The phrase “along the surface” is essential because moving from one surface to another may involve work.
109. At any point on an equipotential surface, the electric field is directed:
ⓐ. Randomly because potential is scalar
ⓑ. Along the surface
ⓒ. Opposite to the surface only for negative charges
ⓓ. Perpendicular to the surface
Correct Answer: Perpendicular to the surface
Explanation: An equipotential surface has the same potential \(V\) at every point on it. If a charge is moved along this surface, the potential difference is zero and no electrostatic work is done. Work by the field would be non-zero if the field had a component along the displacement on the surface. Therefore, the tangential component of \(\vec{E}\) along an equipotential surface must be zero. The only allowed direction of \(\vec{E}\) at the surface is normal to it. Potential is scalar, but its spatial variation still fixes the direction of the electric field.
110. A small positive test charge is placed between two nearby equipotential surfaces marked \(80\,\text{V}\) and \(60\,\text{V}\). The electric field direction is:
ⓐ. Along the \(80\,\text{V}\) surface
ⓑ. From the \(80\,\text{V}\) surface toward the \(60\,\text{V}\) surface
ⓒ. From the \(60\,\text{V}\) surface toward the \(80\,\text{V}\) surface
ⓓ. Along the \(60\,\text{V}\) surface
Correct Answer: From the \(80\,\text{V}\) surface toward the \(60\,\text{V}\) surface
Explanation: Electric field direction is defined as the direction of force on a positive test charge. In electrostatics, a positive charge tends to move in the direction of decreasing potential if released freely. Therefore, \(\vec{E}\) points from higher potential to lower potential. Between the \(80\,\text{V}\) and \(60\,\text{V}\) equipotential surfaces, the direction is from \(80\,\text{V}\) toward \(60\,\text{V}\). The field is also perpendicular to the equipotential surfaces, not along them. The sign convention is tied to a positive test charge, so reversing it would describe the force on a negative charge, not the field direction itself.
111. A two-dimensional drawing shows equipotential lines labelled \(100\,\text{V}\), \(80\,\text{V}\), \(60\,\text{V}\), and \(40\,\text{V}\). The lines are closer together on the left side and farther apart on the right side. The electric field is stronger:
ⓐ. Equally everywhere, because all lines represent potential
ⓑ. Nowhere, because equipotential lines mean \(\vec{E}=0\)
ⓒ. On the left side, where the same \(\Delta V\) occurs over less distance
ⓓ. On the right side, because the equipotential lines are spread farther apart
Correct Answer: On the left side, where the same \(\Delta V\) occurs over less distance
Explanation: The electric field magnitude is related to how rapidly potential changes with distance. If adjacent equipotential lines differ by the same potential value, closer spacing means a larger potential change per metre. A larger potential gradient corresponds to a stronger electric field. Wider spacing means the same change in \(V\) occurs over a larger distance, so the field is weaker. Equipotential lines do not imply zero electric field; they show where potential is constant along each line. The density of equipotential surfaces gives information about field strength.
112. The row that correctly states the relation between equipotential surfaces and electric field is:
| Row | Equipotential feature | Electric field implication |
|---|
| P | Surfaces are closer together | Stronger electric field |
| Q | Surfaces intersect | Two potential values at one point are allowed |
| R | Movement along one surface | Maximum electrostatic work is done |
| S | Field line meets surface | Field line must be tangential to the surface |
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row P
Explanation: When equipotential surfaces are closer together for the same potential difference, the potential changes more rapidly with distance. This means the electric field magnitude is larger there. Row Q is wrong because two different equipotential surfaces cannot assign two values of \(V\) to one point. Row R is wrong because movement along an equipotential surface gives zero electrostatic work. Row S is wrong because \(\vec{E}\) is perpendicular, not tangential, to an equipotential surface. The supported row is the one that connects spacing with field strength.
113. A field line crosses an equipotential surface at an angle of \(90^\circ\). This crossing angle is required because:
ⓐ. Potential becomes a vector at the crossing point
ⓑ. The electric field must be zero on every equipotential surface
ⓒ. The field has no component along the surface
ⓓ. Work is maximum for motion along an equipotential surface
Correct Answer: The field has no component along the surface
Explanation: Along an equipotential surface, the potential difference between nearby points is zero. If the electric field had a tangential component along the surface, it would do work on a charge displaced along that surface. That would contradict the zero potential difference along an equipotential. Therefore, the tangential component of \(\vec{E}\) must vanish. The remaining direction for \(\vec{E}\) is perpendicular to the surface. This does not mean the field itself is zero; it means its direction is normal to the equipotential surface.
114. Assertion: Two equipotential surfaces with different potential values cannot intersect.
Reason: At an intersection point, the electric field would have to be perpendicular to both surfaces.
ⓐ. Assertion false; Reason is true
ⓑ. Both true; Reason explains Assertion
ⓒ. Both true; Reason does not explain
ⓓ. Assertion true; Reason is not true
Correct Answer: Both true; Reason does not explain
Explanation: The Assertion is true because a single point cannot have two different potential values for the same reference choice. If two equipotential surfaces with different \(V\) values intersected, the intersection point would be assigned both values, which is impossible. The Reason can be true in the sense that the field direction must be normal to an equipotential surface. However, the main reason for non-intersection is the uniqueness of potential at a point, not only the perpendicular-field argument. The contradiction in potential value is enough to reject the intersection. The normal-field idea is related but not the most direct explanation of the Assertion.
115. A charge is moved from one equipotential surface at \(120\,\text{V}\) to another at \(90\,\text{V}\). For a positive charge, the electric field does positive work when the movement is:
ⓐ. From \(120\,\text{V}\) to \(90\,\text{V}\)
ⓑ. Along the \(120\,\text{V}\) surface only
ⓒ. From \(90\,\text{V}\) to \(120\,\text{V}\)
ⓓ. Along the \(90\,\text{V}\) surface only
Correct Answer: From \(120\,\text{V}\) to \(90\,\text{V}\)
Explanation: Work done by the electrostatic field for motion from initial point \(1\) to final point \(2\) is \(W_{\text{field}}=q(V_1-V_2)\). For a positive charge, \(q\gt0\), so field work is positive when \(V_1\gt V_2\). Moving from \(120\,\text{V}\) to \(90\,\text{V}\) gives \(V_1-V_2=30\,\text{V}\), so the field work is positive. Moving from lower potential to higher potential would require negative field work for a positive charge. Motion along either equipotential surface alone gives zero work because the potential difference along that surface is zero. This matches the idea that \(\vec{E}\) points from higher potential toward lower potential.
116. In a diagram, equipotential surfaces are labelled \(V\), \(V-10\,\text{V}\), and \(V-20\,\text{V}\) in order from left to right. The field lines should point:
ⓐ. From right to left
ⓑ. Along each equipotential surface
ⓒ. In circles around each equipotential surface
ⓓ. From left to right
Correct Answer: From left to right
Explanation: The labels show that potential decreases from left to right. Electric field points in the direction in which potential decreases most rapidly. Therefore, the field lines should point from the \(V\) surface toward the \(V-20\,\text{V}\) side. Field lines are perpendicular to equipotential surfaces, not along them. Circular field lines would not represent the electrostatic relation between \(\vec{E}\) and \(V\) here. The arrow direction is chosen by decreasing potential for a positive test charge convention.
117. A region has equally spaced, parallel equipotential planes whose values decrease by \(15\,\text{V}\) after every \(0.05\,\text{m}\) along the \(+x\)-direction. The magnitude of the electric field is:
ⓐ. \(3.0\,\text{V m}^{-1}\)
ⓑ. \(0.75\,\text{V m}^{-1}\)
ⓒ. \(300\,\text{V m}^{-1}\)
ⓓ. \(75\,\text{V m}^{-1}\)
Correct Answer: \(300\,\text{V m}^{-1}\)
Explanation: \( \textbf{Given potential change:} \) \(|\Delta V|=15\,\text{V}\).
\( \textbf{Given separation:} \) \(d=0.05\,\text{m}\).
\( \textbf{Required quantity:} \) Magnitude of electric field \(E\).
\( \textbf{Uniform-field relation from equipotential spacing:} \)
\[
E=\frac{|\Delta V|}{d}
\]
\( \textbf{Substitution:} \)
\[
E=\frac{15\,\text{V}}{0.05\,\text{m}}
\]
\( \textbf{Calculation:} \)
\[
E=300\,\text{V m}^{-1}
\]
\( \textbf{Equivalent unit:} \)
\[
1\,\text{V m}^{-1}=1\,\text{N C}^{-1}
\]
\( \textbf{Final answer:} \) The field magnitude is \(300\,\text{V m}^{-1}\).
The sign of the potential change decides direction, while the magnitude uses \(|\Delta V|\) divided by distance.
118. Three statements about equipotential surfaces are listed.
I. \(\vec{E}\) is perpendicular to an equipotential surface.
II. Closer equipotential surfaces indicate a stronger electric field for the same potential difference.
III. A charge moving along an equipotential surface always has maximum change in potential energy.
ⓐ. I and III only
ⓑ. I, II, and III
ⓒ. II and III only
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is true because the electric field has no tangential component along an equipotential surface. Statement II is true because closer surfaces mean a larger potential change per unit distance. Statement III is false because movement along an equipotential surface gives \(\Delta V=0\), so \(\Delta U=q\Delta V=0\). The change in potential energy is not maximum along such a surface; it is zero. This statement set separates direction, spacing, and work ideas. The first two statements describe field geometry, while the third contradicts the meaning of equipotential.
119. The one-dimensional relation between electric field and potential variation along the \(x\)-axis is:
ⓐ. \(E_x=-q\frac{dV}{dx}\)
ⓑ. \(E_x=Vx\)
ⓒ. \(E_x=\frac{dV}{dt}\)
ⓓ. \(E_x=-\frac{dV}{dx}\)
Correct Answer: \(E_x=-\frac{dV}{dx}\)
Explanation: In one dimension, the electric field component along \(x\) is related to the spatial rate of change of potential by \(E_x=-\frac{dV}{dx}\). The derivative must be with respect to position, not time, because electric field is connected to spatial variation of potential. The factor \(q\) is not part of the field-potential relation; multiplying by charge gives force or energy-related quantities. The negative sign shows that the field points in the direction of decreasing potential. A constant potential with no spatial change gives zero field component along that direction. This relation converts a potential-position graph into field information.
120. The \(V\)-versus-\(x\) graph for a region is a straight line with \(V\) decreasing from \(60\,\text{V}\) at \(x=0\) to \(20\,\text{V}\) at \(x=2.0\,\text{m}\). The electric field along \(x\) is:
ⓐ. \(-20\,\text{V m}^{-1}\)
ⓑ. \(+20\,\text{V m}^{-1}\)
ⓒ. \(-40\,\text{V m}^{-1}\)
ⓓ. \(+40\,\text{V m}^{-1}\)
Correct Answer: \(+20\,\text{V m}^{-1}\)
Explanation: \( \textbf{Given graph data:} \) \(V_1=60\,\text{V}\) at \(x_1=0\), and \(V_2=20\,\text{V}\) at \(x_2=2.0\,\text{m}\).
\( \textbf{Slope of the} \) \(V\)-\(x\) \( \textbf{graph:} \)
\[
\frac{dV}{dx}=\frac{V_2-V_1}{x_2-x_1}
\]
\( \textbf{Substitution:} \)
\[
\frac{dV}{dx}=\frac{20\,\text{V}-60\,\text{V}}{2.0\,\text{m}-0}
\]
\( \textbf{Slope value:} \)
\[
\frac{dV}{dx}=\frac{-40\,\text{V}}{2.0\,\text{m}}=-20\,\text{V m}^{-1}
\]
\( \textbf{Field relation:} \)
\[
E_x=-\frac{dV}{dx}
\]
\( \textbf{Field component:} \)
\[
E_x=-(-20\,\text{V m}^{-1})=+20\,\text{V m}^{-1}
\]
\( \textbf{Final answer:} \) The electric field is \(+20\,\text{V m}^{-1}\) along the \(+x\)-direction.
The negative graph slope gives a positive field component because of the minus sign in \(E_x=-\frac{dV}{dx}\).
