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प्रश्न
The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of ______.
विकल्प
1 m
2 m
3 m
6 m
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उत्तर
The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of 3 m.
Explanation:
We know that,
E = `(kq)/r^2`
Where E is the magnitude of the electric field, k is constant, q is a charge of magnitude and r is the distance away from the point charge.
Given: E1 = 9 N/C, r1 = 4 and k = `9 xx 10^9` Nm2/C2, E2 = 16 N/C and q is same for both objects.
To find: r2
`E_1 = (kq)/r_1^2`
`9 = ((9 xx 10^9 xx q))/4^2`
`q = ((9 xx 16))/((9 xx 10^9))`
q = 16 × 10-9
`E_2 = (kq)/r_2^2`
`16 = ((9 xx 10^9 xx 16 xx 10^-9))/r_2^2`
`r_2^2 = 9`
`r_2 = 3` m
संबंधित प्रश्न
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