Question:
Work done in moving a charge on an equipotential surface is:
Work done in moving a charge on an equipotential surface is:
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Since potential remains constant everywhere on an equipotential surface, the potential difference is zero. No potential difference means no work is done.
Additionally, remember that electric field lines and equipotential surfaces are always at a $90^\circ$ angle to each other.
Additionally, remember that electric field lines and equipotential surfaces are always at a $90^\circ$ angle to each other.
Updated On: Jun 3, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the amount of work required to move an electric charge from one point to another on an equipotential surface.
Step 2: Key Formula or Approach:
The work done ($W$) in moving a charge $q$ between two points with a potential difference $\Delta V$ is given by:
\[ W = q \cdot \Delta V = q(V_B - V_A) \]
Step 3: Detailed Explanation:
• An equipotential surface is defined as a surface where every point lies at the exact same electric potential.
• Let us consider two arbitrary points, $A$ and $B$, located on this equipotential surface. By definition, the electric potential at point $A$ ($V_A$) is equal to the electric potential at point $B$ ($V_B$), meaning $V_A = V_B$.
• Consequently, the potential difference ($\Delta V$) between these two points is zero: $\Delta V = V_B - V_A = 0$.
• The electrostatic work done by an external force to move a charge $q$ from $A$ to $B$ is proportional to the potential difference between those two points, expressed as $W = q \cdot \Delta V$.
• Substituting $\Delta V = 0$ into the work formula, we find: $W = q \times 0 = 0$.
• Another way to analyze this is through the relationship between the electric field and the equipotential surface. The electric field lines are always perpendicular to the equipotential surface at every point.
• The electrostatic force acting on the charge is $\vec{F} = q\vec{E}$, which is perpendicular to the surface. Since the movement occurs along the surface, the displacement vector $d\vec{r}$ is perpendicular to the force vector $\vec{F}$.
• Therefore, the work done, which is the line integral of force, becomes zero because the dot product of two perpendicular vectors is zero: $dW = \vec{F} \cdot d\vec{r} = F \cdot dr \cdot \cos(90^\circ) = 0$.
Step 4: Final Answer:
The total work done in moving a charge along an equipotential surface is zero.
The question asks for the amount of work required to move an electric charge from one point to another on an equipotential surface.
Step 2: Key Formula or Approach:
The work done ($W$) in moving a charge $q$ between two points with a potential difference $\Delta V$ is given by:
\[ W = q \cdot \Delta V = q(V_B - V_A) \]
Step 3: Detailed Explanation:
• An equipotential surface is defined as a surface where every point lies at the exact same electric potential.
• Let us consider two arbitrary points, $A$ and $B$, located on this equipotential surface. By definition, the electric potential at point $A$ ($V_A$) is equal to the electric potential at point $B$ ($V_B$), meaning $V_A = V_B$.
• Consequently, the potential difference ($\Delta V$) between these two points is zero: $\Delta V = V_B - V_A = 0$.
• The electrostatic work done by an external force to move a charge $q$ from $A$ to $B$ is proportional to the potential difference between those two points, expressed as $W = q \cdot \Delta V$.
• Substituting $\Delta V = 0$ into the work formula, we find: $W = q \times 0 = 0$.
• Another way to analyze this is through the relationship between the electric field and the equipotential surface. The electric field lines are always perpendicular to the equipotential surface at every point.
• The electrostatic force acting on the charge is $\vec{F} = q\vec{E}$, which is perpendicular to the surface. Since the movement occurs along the surface, the displacement vector $d\vec{r}$ is perpendicular to the force vector $\vec{F}$.
• Therefore, the work done, which is the line integral of force, becomes zero because the dot product of two perpendicular vectors is zero: $dW = \vec{F} \cdot d\vec{r} = F \cdot dr \cdot \cos(90^\circ) = 0$.
Step 4: Final Answer:
The total work done in moving a charge along an equipotential surface is zero.
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