The electric potential at point `A` is `20 V` and at `B` is `-40 V`. Find the work done by an external and electrostatic force in moving an electron slowly from B to A.

Explanation

To find the work done by an external and electrostatic force in moving an electron slowly from point B to point A, we can follow these steps:

### Step 1: Identify the electric potentials at points A and B
- The electric potential at point A (VA) is given as 20 V.
- The electric potential at point B (VB) is given as -40 V.

### Step 2: Calculate the change in electric potential (ΔV)
- The change in electric potential when moving from B to A is calculated as:
\[ \Delta V = V_A - V_B = 20 V - (-40 V) = 20 V + 40 V = 60 V \]

### Step 3: Determine the charge of the electron
- The charge of an electron (q) is approximately:
\[ q = -1.6 \times 10^{-19} \text{ C} \]

### Step 4: Calculate the change in potential energy (ΔPE)
- The change in potential energy when moving the electron from B to A is given by:
\[ \Delta PE = q \cdot \Delta V = (-1.6 \times 10^{-19} \text{ C}) \cdot (60 \text{ V}) \]
- Performing the multiplication:
\[ \Delta PE = -1.6 \times 60 \times 10^{-19} = -96 \times 10^{-19} \text{ J} = -9.6 \times 10^{-18} \text{ J} \]

### Step 5: Calculate the work done by the external force (W_ext)
- The work done by the external force is equal to the change in potential energy:
\[ W_{\text{ext}} = \Delta PE = -9.6 \times 10^{-18} \text{ J} \]

### Step 6: Calculate the work done by the electrostatic force (W_elec)
- The work done by the electrostatic force is the negative of the change in potential energy:
\[ W_{\text{elec}} = -\Delta PE = -(-9.6 \times 10^{-18} \text{ J}) = 9.6 \times 10^{-18} \text{ J} \]

### Final Results
- Work done by external force:
\[ W_{\text{ext}} = -9.6 \times 10^{-18} \text{ J} \]
- Work done by electrostatic force:
\[ W_{\text{elec}} = 9.6 \times 10^{-18} \text{ J} \]