Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The electric force experienced by a charge of 5 × 10 –6 C is 25 × 10 –3 N. Find the magnitude of the electric field at that position of the charge due to the source charges.
Text Solution
Verified by ExpertsThe correct answer is:
A
To find the electric field (E) at the position of the charge, we can use the formula for electric force (F) experienced by a charge (q):
F = qE
We are given:
F = 25 × 10^-3 N
q = 5 × 10^-6 C
We can rearrange the formula to find the electric field:
E = \frac{F}{q}
Substituting the values in:
E = \frac{25 \times 10^{-3} \text{ N}}{5 \times 10^{-6} \text{ C}}
Simplifying the calculation:
E = \frac{25}{5} \times \frac{10^{-3}}{10^{-6}} = 5 \times 10^{3} \text{ N/C}
Therefore, the magnitude of the electric field at that position of the charge is 5000 N/C.
F = qE
We are given:
F = 25 × 10^-3 N
q = 5 × 10^-6 C
We can rearrange the formula to find the electric field:
E = \frac{F}{q}
Substituting the values in:
E = \frac{25 \times 10^{-3} \text{ N}}{5 \times 10^{-6} \text{ C}}
Simplifying the calculation:
E = \frac{25}{5} \times \frac{10^{-3}}{10^{-6}} = 5 \times 10^{3} \text{ N/C}
Therefore, the magnitude of the electric field at that position of the charge is 5000 N/C.
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