The electric field strength 2.9cm from a 10cm diameter metal ball is 7.7×10^4 N/C . What is the charge (in nC) on the ball?
\[\large E = \dfrac{k q}{r^2}\] Rearrange the formula to make q the subject. Then substitute the values you were given in the question to find q. Remember to change all units into SI units.
k is the constant with a value of \[8.99\times10^9 Nm^2C^{-2}\]
Is r the radius of the sphere, the distance from the sphere, or both added together?
I'm not too sure about that part. I'm thinking that it might be the distance from the ball plus the radius of the ball. Not too sure, but if you have the answers, try testing it and see if you're correct.
Ok, here's the answer \[(Er^2)/k=q\] r = .05 + 0.29 \[(77000∗(.05+.029)^2)/k=q\] q = 5.3*10^-8 C q= 53 nC
So was that the correct answer on your answer sheet or do you not have an answer sheet?
Yes, that was the correct answer.
Then well done mate. Excellent work.
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