A charge 10 esu is placed at a distance of 2 cm from a charge 40 esu and 4 cm from another charge -20 esu. The potential energy of the charge 10 esu is : (in ergs)

Options

A. `87.5`
B. `112.5`
C. 150
D. zero

Answer

To find the potential energy of the charge \(10 \, \text{esu}\) in the presence of the other charges, we can use the formula for the potential energy \(U\) between two point charges:

\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]

Where:
- \(U\) is the potential energy,
- \(k\) is the proportionality constant (in electrostatic units, \(k = 1\)),
- \(q_1\) and \(q_2\) are the magnitudes of the charges,
- \(r\) is the distance between the charges.

### Step 1: Calculate the potential energy between charge \(10 \, \text{esu}\) and charge \(40 \, \text{esu}\)

Given:
- \(q_1 = 10 \, \text{esu}\)
- \(q_2 = 40 \, \text{esu}\)
- \(r = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}\)

Using the formula:

\[ U_{1} = \frac{1 \cdot 10 \cdot 40}{2 \times 10^{-2}} = \frac{400}{0.02} = 20000 \, \text{ergs} \]

### Step 2: Calculate the potential energy between charge \(10 \, \text{esu}\) and charge \(-20 \, \text{esu}\)

Given:
- \(q_1 = 10 \, \text{esu}\)
- \(q_2 = -20 \, \text{esu}\)
- \(r = 4 \, \text{cm} = 4 \times 10^{-2} \, \text{m}\)

Using the formula:

\[ U_{2} = \frac{1 \cdot 10 \cdot (-20)}{4 \times 10^{-2}} = \frac{-200}{0.04} = -5000 \, \text{ergs} \]

### Step 3: Calculate the total potential energy

Now, we can find the total potential energy \(U\) of the charge \(10 \, \text{esu}\) by adding the potential energies from both interactions:

\[ U = U_{1} + U_{2} = 20000 \, \text{ergs} + (-5000 \, \text{ergs}) = 15000 \, \text{ergs} \]

### Final Answer

The potential energy of the charge \(10 \, \text{esu}\) is \(15000 \, \text{ergs}\).

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