Question

What is the magnitude of a point charge chosen so that the electric field 50cm away has magnitude 2.0n/c

Answer 1

we know the formula of electric field, it is given by ,

where K is kappa constant e.g., K = 9 × 10^9 Nm²/C² , Q is magnitude of the charge and r is the distance of charge and point of observation.

given, r = 50cm = 0.5m

E = 2N/C

so, 2 = 9 × 10^9Q/(0.5)²

or, 2 × 0.5 × 0.5 = 9 × 10^9 Q

or, 0.5/9 × 10^-9 = Q

or, Q = 0.055 × 10^-9 = 5.5 × 10^-11 C

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