Description
A straight, copper wire has a length of 0.50 m and an excess charge of -1.0 × 10-5 C distributed uniformly along its length. Find the magnitude of the electric field at a point located 7.5 × 10-3 m from the midpoint of the wire.
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Explanation & Answer
charge = q = -1.0 × 10^-5 C
point distance = r = 7.5 × 10^-3 m
electric field = E = ?
Now as we know that
E = k q / r^2
Now putting the values, we get
E = (9 × 10^9) (1.0 × 10^-5 C) / (7.5 × 10^-3)^2
E = 9 × 10^9 - 5 / 56.25 × 10^-6 × 10^2
E = 9 × 10^4 / 56.25 × 10^-4
E = 9 × 10^4 + 4 / 56.25
E = 9 × 10^8 / 56.25
E = 0.16 × 10^8
E = 16 × 10^6
E = 16 G NC^-1
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