A metal sphere with its centre at A and radius R has a charge 2q on it. The field at a point B outside the sphere is E. If another metal sphere of radius 3R and having a charge -3q is placed with its centre at point B, find out the resultant electric field at a point mid way between A and B.

Answer

To find the resultant electric field at a point midway between two metal spheres with given charges, we can follow these steps:

### Step 1: Understand the setup
We have two metal spheres:
- Sphere 1 (centered at point A) has a charge of \(2q\) and radius \(R\).
- Sphere 2 (centered at point B) has a charge of \(-3q\) and radius \(3R\).

We need to find the electric field at point C, which is midway between points A and B.

### Step 2: Determine the distance from A to B
Let the distance between the centers A and B be \(d\). Since point C is midway, the distance from A to C is \(d/2\) and the distance from B to C is also \(d/2\).

### Step 3: Calculate the electric field due to Sphere 1 at point C
The electric field \(E_A\) due to Sphere 1 at point C can be calculated using the formula for the electric field due to a point charge:
\[ E_A = \frac{k \cdot Q}{r^2} \]
where \(Q = 2q\) and \(r = d/2\). Thus,
\[ E_A = \frac{k \cdot 2q}{(d/2)^2} = \frac{k \cdot 2q}{d^2/4} = \frac{8kq}{d^2} \]

### Step 4: Calculate the electric field due to Sphere 2 at point C
The electric field \(E_B\) due to Sphere 2 at point C is given by:
\[ E_B = \frac{k \cdot Q}{r^2} \]
where \(Q = -3q\) and \(r = d/2\). Thus,
\[ E_B = \frac{k \cdot (-3q)}{(d/2)^2} = \frac{k \cdot (-3q)}{d^2/4} = \frac{-12kq}{d^2} \]

### Step 5: Determine the resultant electric field at point C
The resultant electric field \(E_C\) at point C is the vector sum of \(E_A\) and \(E_B\):
\[ E_C = E_A + E_B = \frac{8kq}{d^2} + \frac{-12kq}{d^2} = \frac{-4kq}{d^2} \]

### Step 6: Conclusion
The resultant electric field at point C is:
\[ E_C = \frac{-4kq}{d^2} \]
This indicates that the electric field at point C is directed towards Sphere 2, due to the negative charge on it.

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