The electric field at a point due to a point charge is 30 N/C and the electric potential at that point is 15 J/C respectively. What is the distance of the point from the charge and the magnitude of the charge?0.1 m, 0.43 nC0.4 m, 0.77 nC0.5 m, 0.83 nC0.6 m, 0.31 nC
Question
The electric field at a point due to a point charge is 30 N/C and the electric potential at that point is 15 J/C respectively. What is the distance of the point from the charge and the magnitude of the charge?0.1 m, 0.43 nC0.4 m, 0.77 nC0.5 m, 0.83 nC0.6 m, 0.31 nC
Solution
The electric field E at a point due to a point charge is given by the formula:
E = kQ/r^2
where:
- E is the electric field,
- k is Coulomb's constant (approximately 8.99 x 10^9 N m^2/C^2),
- Q is the charge, and
- r is the distance from the charge to the point.
The electric potential V at a point due to a point charge is given by the formula:
V = kQ/r
where:
- V is the electric potential,
- k is Coulomb's constant,
- Q is the charge, and
- r is the distance from the charge to the point.
Given that E = 30 N/C and V = 15 J/C, we can set up the following system of equations:
30 = kQ/r^2 15 = kQ/r
Dividing the first equation by the second, we get:
30/15 = (kQ/r^2) / (kQ/r) 2 = r
So, the distance of the point from the charge is 0.5 m.
Substituting r = 0.5 m into the second equation, we get:
15 = kQ/0.5 Q = 15 * 0.5 / k Q = 7.5 / k
Substituting k = 8.99 x 10^9 N m^2/C^2, we get:
Q = 7.5 / (8.99 x 10^9) Q = 0.83 x 10^-9 C
So, the magnitude of the charge is 0.83 nC.
Therefore, the correct answer is 0.5 m, 0.83 nC.
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