What must the magnitude of an isolated positive point charge be for the electric potential at 9 cm from the charge be +170 V? Give your answer in coulombs to 3 significant figures in scientific notation
Question
What must the magnitude of an isolated positive point charge be for the electric potential at 9 cm from the charge be +170 V? Give your answer in coulombs to 3 significant figures in scientific notation
Solution
The formula for the electric potential (V) due to a point charge is given by:
V = k * (Q/r)
where: V is the electric potential, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
We are given V and r, and we are asked to solve for Q. Rearranging the formula to solve for Q gives:
Q = V * r / k
First, we need to convert the given distance to the correct units:
r = 9 cm = 9 x 10^-2 m = 0.09 m
Now we can substitute the given values into the formula:
Q = (170 V * 0.09 m) / (8.99 x 10^9 N m^2/C^2) Q = 1.70 x 10^-9 C
So, the magnitude of the charge is approximately 1.70 x 10^-9 Coulombs, or 1.70 nC. In scientific notation to 3 significant figures, this is 1.70 x 10^-9 C.
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