Equipotential Curves for Given Electric Field – Rankers Physics

Electric Potential: Practice Problem & Solution

Assertion (A): If electric field in x-y plane is given by \( \vec{E} = y \hat{i} + x \hat{j} \) then equipotential curve is given by \( xy = \text{constant} \). Reason (R): Electric field may not be perpendicular to equipotential surface/curve/line.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Electric Potential. Understanding the underlying formula is key to arriving at the correct answer below:

For \( \vec{E} = y \hat{i} + x \hat{j} \), we have \( dV = -\vec{E} \cdot d\vec{l} = -(y dx + x dy) = -d(xy) \). Integrating, \( V = -xy + C \). Thus, equipotential lines are \( xy = \text{constant} \). Electric field lines are always perpendicular to equipotential surfaces. Thus (A) is true and (R) is false.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A hollow sphere of radius $1 \text{ m}$ is given a positive charge of $10 \mu \text{C}$. The electric field at the centre of hollow sphere will be: (1990)
2. A large sphere \( P \) of radius \( R \) is charged positively. It is momentarily connected to a small sphere \( Q \) of radius \( r \). The two spheres now have same
3. In a certain region of space, the potential is given by : V = k[2x² + y² – 2z]. The electric field at the point (1, 0, 1) has magnitude equal to 
4. If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$, the electric field (in N/C) at point $(1, 1, 0)$ is: (2015)
5. According to Gauss's law in electrostatics, net electric flux through a closed surface depends on