ELECTROSTATICS O13

Electrostatic Force Solution

Problem: Force between Half-Cylinder and Plate

Given: A half cylinder (radius $R$, length $L$) with charge $Q$, and a base plate (width $2R$, length $L$) with charge $q$. We need to find the electrostatic force between them.

Step 1:

The force on the plate can be calculated using the field created by the cylinder. Since the setup is symmetric, the net force on the plate will be only due to the perpendicular component of electric field.

Therefore, we can use the relation:

$$F = \int E \cdot dq cos\theta= \sigma_{\text{plate}} \int E \cdot dA = \sigma_{\text{plate}} \cdot \Phi$$

Where $\sigma_{\text{plate}}$ is the charge density of the plate and $\Phi$ is the electric flux of the cylinder passing through the plate.

Step 2: Calculate Charge Density ($\sigma$)

The charge $q$ is uniformly distributed over the rectangular plate of area $2R \times L$.

$$\sigma_{\text{plate}} = \frac{q}{\text{Area}} = \frac{q}{2RL}$$

Step 3: Calculate Flux ($\Phi$)

This is the trickiest part. Consider the cross-section (see diagram):

Any point charge $dQ$ on the semi-circle “sees” the diameter (the plate).
From geometry, the angle subtended by the diameter at any point on the circumference is always $90^\circ$ ($\pi/2$).
Since the total angle around a point in 2D is $360^\circ$ ($2\pi$), the fraction of flux going through the plate is:

$$\text{Fraction} = \frac{90^\circ}{360^\circ} = \frac{1}{4}$$

So, the total flux through the plate is $1/4$ of the total flux emitted by $Q$.

$$\Phi = \frac{1}{4} \frac{Q}{\epsilon_0}$$

Step 4: Final Calculation

Substitute $\sigma$ and $\Phi$ into the force equation:

$$F = \sigma_{\text{plate}} \times \Phi$$ $$F = \left( \frac{q}{2RL} \right) \times \left( \frac{Q}{4\epsilon_0} \right)$$ $$F = \frac{qQ}{8\epsilon_0 RL}$$

Answer: (d) $\frac{qQ}{8\epsilon_0 RL}$