A particle having a charge`+ 3 xx 10^(-9 )C` is placed in a uniform electric field directed toward left. It is released from rest and moves a distance of 5 cm after which its kinetic energy is found to be `4.5 xx 10^(-5) J`.
(a) Calcnalate the work done by the electrical force on the particle
(b) Calculate the magnitude of the electric field.
(c) Calculate the potential of starting point with respect to the end point of particle's motion.

Answer

Let's solve the problem step by step.

### Given Data:
- Charge of the particle, \( q = +3 \times 10^{-9} \, \text{C} \)
- Distance moved, \( d = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
- Kinetic energy after moving the distance, \( KE = 4.5 \times 10^{-5} \, \text{J} \)

### Part (a): Calculate the work done by the electrical force on the particle.

According to the work-energy principle, the work done \( W \) by the electric force on the particle is equal to the change in kinetic energy:

\[ W = KE_{\text{final}} - KE_{\text{initial}} \]

Since the particle starts from rest, its initial kinetic energy \( KE_{\text{initial}} = 0 \).

Thus, the work done is:

\[ W = KE_{\text{final}} - 0 = KE_{\text{final}} = 4.5 \times 10^{-5} \, \text{J} \]

### Part (b): Calculate the magnitude of the electric field.

The work done by the electric force can also be expressed as:

\[ W = F \cdot d \]

Where \( F \) is the force experienced by the charge in the electric field, given by:

\[ F = qE \]

Substituting this into the work equation:

\[ W = qE \cdot d \]

Now, we can solve for the electric field \( E \):

\[ E = \frac{W}{qd} \]

Substituting the known values:

\[ E = \frac{4.5 \times 10^{-5} \, \text{J}}{(3 \times 10^{-9} \, \text{C})(5 \times 10^{-2} \, \text{m})} \]

Calculating the denominator:

\[ (3 \times 10^{-9})(5 \times 10^{-2}) = 15 \times 10^{-11} \, \text{C}\cdot\text{m} \]

Now substituting back into the equation for \( E \):

\[ E = \frac{4.5 \times 10^{-5}}{15 \times 10^{-11}} = 3 \times 10^{6} \, \text{N/C} \]

### Part (c): Calculate the potential of the starting point with respect to the end point of the particle's motion.

The electric potential difference \( V \) between the starting point and the endpoint can be calculated using the formula:

\[ V = \frac{W}{q} \]

Substituting the known values:

\[ V = \frac{4.5 \times 10^{-5} \, \text{J}}{3 \times 10^{-9} \, \text{C}} = 15 \times 10^{3} \, \text{V} = 15 \, \text{kV} \]

### Summary of Answers:
(a) Work done by the electrical force: \( 4.5 \times 10^{-5} \, \text{J} \)

(b) Magnitude of the electric field: \( 3 \times 10^{6} \, \text{N/C} \)

(c) Potential of starting point with respect to the end point: \( 15 \, \text{kV} \)