Question

An infinite charged plane with charge density $(2\varepsilon_0)$ C/m² is tilted at angle 53° with vertical about point O as shown. There are two points $P$ and $Q$ at separation 2.5 m on the vertical line passing through point O. The potential different between point $P$ and $Q$, $|V_P - V_Q|$ in volt is

Answer 1

2.0

Answer 2

The problem asks for the potential difference between two points P and Q due to an infinite charged plane.

1. Electric Field of an Infinite Charged Plane: The electric field produced by an infinite plane with uniform surface charge density $\sigma$ is uniform and perpendicular to the plane. Its magnitude is given by: $E = \frac{\sigma}{2\varepsilon_0}$

2. Calculate the Magnitude of the Electric Field: Given the charge density $\sigma = (2\varepsilon_0)$ C/m². Substituting this value into the formula: $E = \frac{2\varepsilon_0}{2\varepsilon_0} = 1 \text{ V/m}$

3. Relate Potential Difference to Electric Field: The potential difference between two points P and Q is given by the line integral of the electric field: $V_P - V_Q = -\int_Q^P \vec{E} \cdot d\vec{l}$ For a uniform electric field, this simplifies to: $V_P - V_Q = - \vec{E} \cdot \vec{r}_{QP}$ where $\vec{r}_{QP}$ is the displacement vector from point Q to point P. The magnitude of this potential difference is $|V_P - V_Q| = |E d \cos\theta|$, where $d$ is the magnitude of the displacement vector $\vec{r}_{QP}$ and $\theta$ is the angle between the electric field vector $\vec{E}$ and the displacement vector $\vec{r}_{QP}$.

4. Analyze the Geometry and Determine the Angle:

   • The separation between points P and Q is $d = 2.5$ m.
   • Points P and Q are on a vertical line. Let's assume the displacement vector $\vec{r}_{QP}$ points vertically upwards.
   • The infinite charged plane is tilted at an angle of 53° with the vertical.
   • The electric field $\vec{E}$ is always perpendicular to the plane.
   • Therefore, the angle between the electric field vector $\vec{E}$ and the vertical line (along which P and Q lie) is $90^\circ - 53^\circ = 37^\circ$. So, $\theta = 37^\circ$.

5. Calculate the Potential Difference: We have $E = 1$ V/m, $d = 2.5$ m, and $\theta = 37^\circ$. We use the approximation for trigonometric values: $\cos(37^\circ) \approx \frac{4}{5} = 0.8$.

   $V_P - V_Q = - E d \cos\theta$ $V_P - V_Q = - (1 \text{ V/m}) (2.5 \text{ m}) \cos(37^\circ)$ $V_P - V_Q = - (1) (2.5) (0.8)$ $V_P - V_Q = - 2.0 \text{ V}$

6. Find the Absolute Potential Difference: The question asks for $|V_P - V_Q|$. $|V_P - V_Q| = |-2.0 \text{ V}| = 2.0 \text{ V}$