A charge moves a distance of 1.4 cm in the
direction of a uniform electric field having
a magnitude of 244 N/C. The electrical
potential energy of the charge decreases by
108.3 × 10−19 J as it moves.
Find the magnitude of the charge on the
moving particle. The electrical potential energy depends on the distance moved in the
direction of the field.
Answer in units of C
To find the magnitude of the charge on the moving particle, we can use the relationship between electric potential energy and the amount of work done on the charge.
The electric potential energy (PE) of a charge (q) moving in an electric field (E) is given by the formula:
PE = qE*d
Where:
PE is the electrical potential energy
q is the magnitude of the charge
E is the magnitude of the electric field
d is the distance moved in the direction of the field
In this problem, we are given the following values:
PE = -108.3 × 10^(-19) J (Note that the negative sign indicates a decrease in potential energy)
E = 244 N/C
d = 1.4 cm = 0.014 m
Substituting these values into the formula, we get:
-108.3 × 10^(-19) J = q * 244 N/C * 0.014 m
Now we can solve for q:
q = (-108.3 × 10^(-19) J) / (244 N/C * 0.014 m)
Calculating this expression, we find:
q ≈ -1.342 × 10^(-19) C
Since charge cannot be negative, we take the magnitude:
|q| ≈ 1.342 × 10^(-19) C
Therefore, the magnitude of the charge on the moving particle is approximately 1.342 × 10^(-19) C.