Q.A point charge $+q$ is placed at a distance $d$ from an isolated conducting plane. The field at a point $P$ on the other side of the plane is

At a conductor's surface the electrostatic field must be perpendicular to the surface. The isolated plate is polarised by $+q$: the near face gains negative induced charge and the far face gains positive charge, so just outside the far side the field is normal to the plate and points outward (away from it). The radial options (c) and (d) ignore …

Concept understanding. A conductor in electrostatic equilibrium has zero field inside, and at its surface the field is always normal — any tangential component would drive the free charges until it vanished. The external charge $+q$ polarises the plate: the near face becomes negatively charged, and because the plate is isolated (net charge zero) the far face becomes positively charged.

Field at $P$ on the far side. Point $P$ lies just outside the far surface, so the resultant field there is perpendicular to the plane. The positive induced charge on that face produces a field directed outward, i.e. away from the plate. It is therefore neither radial from $q$ nor directed toward the plate.

Checking the options. …

Concept: Method of Images for a Point Charge Near a Conducting Plane

The Method of Images replaces the conducting plane with an image charge of opposite sign placed symmetrically on the other side. This lets you calculate the field without solving boundary conditions directly.

Method: Method of Images

Steps:

1. Remove the conducting plane and replace it with an image charge $q' = -q$ placed at a distance $d$ behind the plane's original position (i.e., at a perpendicular distance $d$ on the opposite side).

2. Identify the point $P$ where the field is to be found. It lies on the other side of the plane from the original charge $+q$.

3. Calculate the electric field at $P$ due to both the real charge $+q$ and the image charge $-q$, using Coulomb's law for point charges:

   • Field due to $+q$:

     $\vec{E}_1 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r_1^2} \, \hat{r}_1$

     where $r_1$ is the distance from $+q$ to $P$.

   • Field due to $-q$:

     $\vec{E}_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(-q)}{r_2^2} \, \hat{r}_2$

     where $r_2$ is the distance from the image charge to $P$.

4. Add the vectors (superposition): …

Here are the common mistakes students make when analyzing the field due to a point charge near an isolated conducting plane, along with how to avoid each.

1. Forgetting the Method of Images

Mistake:

Students try to calculate the field at point $P$ by considering only the real charge $+q$ and the induced charges on the plane. They attempt to integrate over the induced charge distribution, which is messy and often leads to errors.

Why it’s wrong:

The induced charge distribution is not uniform — it’s denser near the point closest to $+q$. Direct integration is unnecessarily complex and error-prone.

How to avoid:

Use the method of images. Replace the conducting plane with an image charge $-q$ placed symmetrically on the opposite side of the plane (at distance $d$ behind it). The field on the other side of the plane (the side without the real charge) is then simply the field due to the image charge alone (since the real charge’s field is zero inside the conductor).

Key rule: For an isolated conducting plane, the field on the opposite side is exactly the field of the image charge.

2. Confusing “Other Side” with “Same Side”

Mistake:

Students calculate the field at $P$ as the sum of fields from both $+q$ and the image $-q$.

Why it’s wrong:

The method of images gives the correct field only on the side of the real charge. On the opposite side (where $P$ is located), the real charge’s field is zero inside the conductor, and the field outside is due entirely to the induced charges — which is equivalent to the image charge alone.

How to avoid:

Draw a clear diagram. Mark which side of the plane contains $+q$ and which side contains $P$.

• If $P$ is on the same side as $+q$: field = superposition of $+q$ and $-q$ (image).
• If $P$ is on the opposite side: field = field of only the image charge $-q$.

3. Ignoring the “Isolated” Condition

Mistake:

Treating the plane as grounded (connected to Earth) instead of isolated.

Why it’s wrong:

• For a grounded plane, the total induced charge is $-q$, and the image charge is $-q$.
• For an isolated plane, the plane has zero net charge. The induced charge on the near side is $-q$, but an equal $+q$ appears on the far side (or on the edges). This changes the field far away.

How to avoid:

Check the problem statement:

• Grounded plane → image charge = $-q$ at distance $d$ behind.
• Isolated plane → you must add a second image charge $+q$ at the same location as the first image (or at infinity) to keep net charge zero. For points very close to the plane, the effect of the $+q$ is negligible, but for distant points it matters.

Exam tip: If the problem says “isolated conducting plane,” the field on the opposite side is not simply the field of $-q$ — it’s the field of $-q$ plus a uniform field from the $+q$ distributed on the far side.

4. Misplacing the Image Charge

Mistake:

Placing the image charge at the wrong distance or on the wrong side.

Why it’s wrong:

The method of images requires the image to be symmetrically placed with respect to the plane. If $+q$ is at distance $d$ from the plane, the image $-q$ must be at distance $d$ behind the plane (on the opposite side).

How to avoid: …