A circle of radius \(r\) is drawn in a uniform electric field \(E\) as shown in figure. If \(V_A, V_B, V_C\) and \(V_D\) are the potentials at \(A,B,C\) and \(D\) respectively, then:
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In uniform electric field:
\[
V=-Ex
\]
Potential depends only on displacement along field direction.
Step 1: Link potential to position. In a uniform field pointing along the horizontal, the potential only depends on the horizontal position $x$. \[ V = -E x \] So points at the same $x$ have the same potential.
Step 2: Look at points A and B. In the figure $A$ and $B$ lie one above the other, so they share the same horizontal position. Step 3: Compare A and B. Same $x$ means same potential. \[ V_A = V_B \] Step 4: Look at points C and D. These lie along the horizontal direction. Potential drops as we move in the direction the field points. Step 5: Compare C and D. Point $D$ is on the higher potential side and $C$ on the lower side, so \[ V_D > V_C \quad\text{that is}\quad V_C < V_D \] Step 6: Combine both results. \[ \boxed{V_A = V_B,\; V_C < V_D} \]
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