A ball with charge `-50 e` is placed at the centre of a hollow spherical shell has a net charge of `-50e`. What is the charge on the shell's outer surface ?

Options

A. `-50 e`
B. zero
C. `-100e`
D. `+100 e`

Answer

To solve the problem, we need to analyze the situation involving the charged ball and the hollow spherical shell. Here is the step-by-step solution:

### Step 1: Understand the setup
We have a ball with a charge of \(-50e\) placed at the center of a hollow spherical shell. The shell itself has a net charge of \(-50e\).

### Step 2: Apply Gauss's Law
According to Gauss's Law, the electric field inside a conductor in electrostatic equilibrium is zero. This means that any excess charge on the shell will redistribute itself in such a way that the electric field inside the shell (and thus at the location of the charge) is zero.

### Step 3: Determine the induced charge on the inner surface of the shell
The charge on the ball \(-50e\) will induce a charge of \(+50e\) on the inner surface of the shell. This is because the inner surface must have a charge that cancels the electric field due to the charge at the center to maintain zero electric field inside the conductor.

### Step 4: Calculate the charge on the outer surface of the shell
The total charge on the shell is \(-50e\). Since we have induced \(+50e\) on the inner surface, we can find the charge on the outer surface by using the equation:

\[ \text{Charge on outer surface} = \text{Total charge on shell} - \text{Charge on inner surface} \]

Substituting the values:

\[ \text{Charge on outer surface} = -50e - (+50e) = -50e - 50e = -100e \]

### Conclusion
The charge on the outer surface of the shell is \(-100e\).

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