An 85nC of source charge generate an electric potential of 115 J/C against a test charge. How far away is the source charge?Group of answer choices6.7 m6.7 cm1.2x10^19 m7 cm
Question
An 85nC of source charge generate an electric potential of 115 J/C against a test charge. How far away is the source charge?Group of answer choices6.7 m6.7 cm1.2x10^19 m7 cm
Solution
The formula for electric potential (V) is given by:
V = kQ/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 source charge, and
- r is the distance from the source charge.
We are given that V = 115 J/C and Q = 85 nC = 85 x 10^-9 C. We are asked to find r.
Rearranging the formula for r gives:
r = kQ/V
Substituting the given values gives:
r = (8.99 x 10^9 N m^2/C^2 * 85 x 10^-9 C) / 115 J/C
Solving this gives r ≈ 6.7 m. So, the source charge is approximately 6.7 meters away.
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