How Is Charge q2 Calculated Using Electric Potential and Field Equations?

AI Thread Summary
To calculate charge q2 using electric potential and field equations, the relationship between electric field and potential difference must be understood. The equation E = kQ/r^2 is relevant, but the potential difference of 3.2 kV must also be incorporated into the calculations. The initial attempt to solve for q2 yielded an incorrect value of 1.28 µC, highlighting the need to account for the potential difference. The correct value of q2 is 3.5 nC, emphasizing the importance of using all given data in the calculations. Understanding the connection between electric potential and electric field is crucial for solving such problems.
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Have exam tomorrow. I know I should have studied before and more, but please help. If you know the answer, please take some time to reply. Thanks very much.

Homework Statement



There are two charges, q1 = +4.5nC, and q2, ?, which are at distances d1 = 2.5cm and d2 = 2.0cm, from a point which has the tension of 3.2kV. What is the value of q2?

Homework Equations



E = \frac{kQ}{r^2}, I think that's all, but I most probably am wrong.

The Attempt at a Solution



This is certainly wrong. I have not even used one of the data given. Anyone, that's the best I was able to come up with.

\frac{kq_1}{r_1^2} = \frac{kq_2}{r_2^2}

which gives q_2 = 1.28 * 10^{-6} which is a wrong answer, since the correct one is supposed to be 3.5nC.
 
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You have not used that the potential difference U is 3.2 kV.
What does potential have to do with (difference in) electric field E?
 
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