Question

A uniform electric field pointing in the positive X direction exists in a region. Let O be the origin, A be the point on the X-axis at r = +2 cm, and B be the point on the Y-axis at y = +1 cm. Then the potential at the points O, A, and B satisfy:

• A. V_O​>V_A
• B. V_O​<V_A​
• C. V_O​>V_B
• D. V_O​<V_B

Answer 1

The Correct Option is A

In a uniform electric field \(E\), the potential difference \(V\) between two points in the field can be calculated using the equation \(V = E \cdot d\), where \(d\) is the distance in the direction of the field. 

Given the electric field points in the positive X direction, let's evaluate potential differences:

• At Point O (Origin): Assume zero potential, \(V_O = 0\).
• At Point A (r = +2 cm on X-axis): Since the electric field is along the X-axis, \(d = 2 \, \text{cm}\) (convert to meters, \(0.02 \, \text{m}\)).
  The potential \(V_A = -E \cdot 0.02\) because distance in electric field direction decreases potential.
• At Point B (y = +1 cm on Y-axis): Since there's no movement along the field direction, there is no change in potential from origin. Therefore, \(V_B = V_O = 0\).

Comparing potentials:

• \(V_O = 0\) and \(V_A = -E \cdot 0.02\), so \(V_O > V_A\).
• \(V_O = V_B = 0\), so \(V_O = V_B\).

Thus, the correct answer is \(V_O > V_A\).

Answer 2

The electric potential decreases in the direction of the electric field. Since the electric field points along the positive X direction, it means that as you move in this direction, the potential decreases.

Point O is at the origin, so it’s the reference point. Point A is along the X-axis at r=+2 cm, which is in the direction of the field.

 Therefore, the potential at A will be lower than at O. Point B is along the Y-axis at y=+1 cm. 

Since there is no component of the electric field along the Y-axis, no work is done in moving a charge from O to B. 

Hence, the potential at B is the same as at O.

Thus, we have V_O​=V_B​ and V_O​\(>\)V_A​, which means the correct answer is (1).