Recent content by rnjscksdyd

Sorry for late reply, my internet was down -_-;; this country :( :( what do you mean when it isn't (d+x)? Does that mean... when that value is negative, the field will be 0 on the left side and on the right if the value is positive?

I get it now! :D thank you so much ;D

Sorry, forgot about the squares :D Then, is this right? \frac{-7μC}{(d+x)^2}= \frac{+2μC}{x^2} -7μCx^2 = +2μC(d+x)^2 -7μCx^2 = +2μC(d^2 + 2dx + x^2) -7x^2 = 2d^2 + 4dx + 2x^2 I took out μC for now (it looks better :D) 0 = 9x^2 + 4dx + 2d^2 x=\frac{-4d \pm \sqrt{16d^2 - 72d^2}}{18}...

Homework Statement Venus and Earth may be regarded as behaving as black bodies. The mean temperature at the surface of Venus is about 600K and at the surface of Earth is about 300K. Which of the following is the best estimate for the ratio...

I thought 'r' was a distance between the two charges. So, 'r' should be different in that, say, 'r1' is the distance between -7μC and the point where electric field strength is 0 and 'r2' is the distance between +2μC and the point where electric field strength is 0, right?? Could you explain...

Homework Statement Two isolated point charges, -7 μC and +2 μC, are at a fixed distance apart. At which point is it possible for the electric field strength to be zero? (A)......(B)...(C).....(D) o----------------------o----------o---------------o ......-7 μC...+2 μC...