What is the electric field strength at a point in space 4 m due east of a particle with a charge of +8 μC? What
is the direction of the field here?
Answer
Electric field at a point in space can be given as
E=kqr2E=\frac{kq}{r^2}E=r2kq
So
E=(9×109)(8×10−6)42E=4500N/CE=\frac{(9\times10^9) (8\times10^{-6}) }{4^2}\\E=4500N/CE=42(9×109)(8×10−6)E=4500N/C
Direction will be towards east.
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