Question

Two particles are fixed to an x axis: particle 1 of charge -1.00 x 10-7 C is at the origin and particle 2 of charge +1.00 x 10-7 C is at x = 17.1 cm. Midway between the particles, what is the magnitude of the net electric field?

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Answer 1

The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.

The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = -7.041 N/C

Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*7.041 N/C = 14.082 N/C