Two point charges of -40.0 x 10^-6 C and 50.0 x 10^-6 C are placed 12.0 cm apart in air. What is the electric field intensity at a point midway between them?
My answer was 2.25 x 10^8 N/C, however, the back of my textbook says that the answer should be 2.25 x 10^14 N/C. Can anyone verify either answer? Help appreciated!
My answer was 2.25 x 10^8 N/C, however, the back of my textbook says that the answer should be 2.25 x 10^14 N/C. Can anyone verify either answer? Help appreciated!
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Your answer is correct.
E = kq/r + kQ/r
= [(9x10^9)(40x10^-6)/(.0036)] + [(9x10^9)(50x10^-6)/(.0036)]
= 1 x 10^8 + 1.25 x10^8
= 2.25 x 10^8
E = kq/r + kQ/r
= [(9x10^9)(40x10^-6)/(.0036)] + [(9x10^9)(50x10^-6)/(.0036)]
= 1 x 10^8 + 1.25 x10^8
= 2.25 x 10^8