Question

Two positive charges Q and 2Q are kept at points A and B, separated by a distance 2d, as shown in the figure. MCL is a semicircle of radius 2d centered at the origin O. If Q=2C and d=10cm, the value of the line integral \(∫^L_M{\overrightarrow{E}}.{\overrightarrow{dl}}\) (where \(\overrightarrow{E}\) represents electric field) along the path MCL will be____V.

Answer 1

Correct Answer: 0

Line Integral of the Electric Field and Electric Potential 

The line integral of the electric field along a path is related to the change in electric potential \( V \) between the start and end points of the path. Mathematically:

\[ \oint_{\mathcal{C}} \mathbf{E} \cdot d\mathbf{l} = V_C - V_M \]

where:

• \( V_C \) and \( V_M \) are the electric potentials at points C and M, respectively.

1. Electric Potential at a Point

The electric potential due to a point charge \( q \) at a distance \( r \) is given by:

\[ V = \frac{kq}{r} \]

where \( k \) is Coulomb’s constant.

2. Potential at Points C and M

Both points \( C \) and \( M \) are at the same distance \( 2d \) from the charges \( Q \) (at A) and \( 2Q \) (at B), since \( MCL \) is a semicircle of radius \( 2d \). Therefore, the electric potential at both \( C \) and \( M \) is the same.

For any point on the semicircle:

\[ V = \frac{kQ}{2d} + \frac{k(2Q)}{2d} = \frac{3kQ}{2d} \]

Thus,

\[ V_C = V_M \]

3. Line Integral

Since the potentials at points \( C \) and \( M \) are equal (\( V_C = V_M \)):

\[ \oint_{\mathcal{C}} \mathbf{E} \cdot d\mathbf{l} = V_C - V_M = 0 \]

Final Answer

The line integral of the electric field along the given path is zero.