Question

Suppose that the electric field experienced due to a positively charged small spherical conductor at a certain distance is E. What will be the percentage change in electric field experienced at thrice the distance if the charge on the conductor is doubled?

Final Answer

$-77.8 \%$

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Video Transcript

This is College Physics Answers with Shaun Dychko.
The electric field in the first scenario is Coulomb's constant times the charge one, divided by the original distance from the charge which we call

*r one*squared. Then the second scenario we'll call it electric field two is*k Q two*over*r two*squared. We're told that*R two*is three times*r one*. So the distances tripled. I sue an old-fashioned word "thrice" to mean times three. So*r two*is three times*r one*and*Q two*is going to be double or times two*Q one*. So we can substitute each of these in this*E two*formula. So we'll replace*Q two*with two*Q one*and we'll replace*r two*with three*r one*. That three is in brackets here, also squared. So we end up with two*k Q one*over nine*r one*squared.*k Q one*over*r one*squared is*E one*. So we can replace all that with*E one*. So*E two*is two over nine times*E one*. Now, the question is asking for the percent change. Here we go, percent change. So that means you take the difference between*E two*and*E one*and then divide by*E one*. So that's two ninths*E one*, that's what*E two*is, minus*E one*E one, times 100 percent. Factor out the*E one*and you end up with*E one*times two ninths minus one in brackets, and then the*E ones*cancel and we're left with this line here, two ninths minus one, times 100 percent. So that's negative seven ninths because this one can be written as nine over nine. Then two minus nine is seven all over nine, negative seven I should say. This works out to negative 77.8 percent. So there is a 77.8 percent decrease in the electric field when tripling the distance and doubling the charge.