Question

As shown in the figure, if the values of the electric potential at three points A, B and C in a uniform electric field (\( \vec{E} \)) are \(V_A\), \(V_B\), and \(V_C\) respectively, then

• A. \( V_A>V_B>V_C \)
• B. \( V_A>V_C>V_B \)
• C. \( V_C>V_B>V_A \)
  % The options in image are different. Using textual analysis: % Image description suggests B is to the left of A and C. A and C are vertically aligned. E points right. % Potential decreases in direction of E. So B (most left) has highest potential. A and C are on equipotential line. % Let's use the actual figure if possible or standard conventions. % My verbal description implies: x_B<x_A = x_C. % If E is along +x, V decreases with x. So V_B>V_A = V_C. This is not among options typically. % Let's assume the typical diagram where A, B, C are distinct, E points right. % C---A % | / % B % The options are: VA>VB>VC, VA>VC>VB, VC>VB>VA, VC>VA>VB. % The image shows points: % .C .A --→ E (field lines pointing right) % .B % In this configuration: A and C are on a vertical line. B is to the left and slightly below A. % Equipotential lines are perpendicular to E-field lines. So equipotential lines are vertical. % Potential decreases in the direction of E. % x-coordinate of C = x-coordinate of A. So \(V_C = V_A\). % x-coordinate of B is to the left of A and C. So \(x_B<x_A = x_C\). % Since E points right (increasing x), potential decreases as x increases. % So, \(V_B>V_A\) and \(V_B>V_C\). And \(V_A = V_C\). % Thus \(V_B>V_A = V_C\). % This is not in the options. Let's look at the provided options in text:
• D. \( V_A>V_B>V_C \)

Hint:

- Electric field lines point from higher potential to lower potential. - The potential decreases as you move in the direction of the electric field. - Equipotential surfaces are always perpendicular to the electric field lines. In a uniform field, these are parallel planes. - If \( \vec{E} \) points along the +x axis, then \(V\) decreases as \(x\) increases. So if \(x_1<x_2\), then \(V(x_1)>V(x_2)\).

Answer 1

The Correct Option is C

In a uniform electric field \( \vec{E} \), the electric potential \(V\) decreases in the direction of \( \vec{E} \).
Equipotential surfaces are perpendicular to the electric field lines.
The figure (interpreted from typical representations for such options) shows the electric field \( \vec{E} \) pointing from left to right.
Option (3) states \( V_C>V_B>V_A \).
For this to be true, if \( \vec{E} \) points in the positive x-direction, then point C must have the smallest x-coordinate, followed by B, and then A having the largest x-coordinate.
That is, \( x_C<x_B<x_A \).
Points further to the left (opposite to the direction of \( \vec{E} \)) will be at a higher potential.
Points further to the right (in the direction of \( \vec{E} \)) will be at a lower potential.
So, if the points are arranged horizontally such that C is to the left of B, and B is to the left of A, with \( \vec{E} \) pointing right, then \( V_C>V_B>V_A \).
The diagram shown in the problem (with C and A vertically aligned and B to their left, E pointing right) leads to \(V_B>V_A = V_C\), which is not among options.
Assuming the configuration that leads to option (3): C is most "upstream" in the field, A is most "downstream", B is in between.
\( x_C<x_B<x_A \).
E field points from left to right.
Then \( V_C \) is highest, \( V_A \) is lowest, \( V_B \) is in between.
So \( V_C>V_B>V_A \).