Two identical negative charges Q are placed on equal distance r of a positive charge Qb (see photo). What is then the value of Q causing the charges (relative to each other) to be in balance? Thank you!

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Two identical negative charges Q are placed on equal distance r of a positive charge Qb (see photo). What is then the value of Q causing the charges (relative to each other) to be in balance? Thank you!

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Answer 1 · 2017-11-03T23:48:05

The result is $Q = 4 \cdot Q_{b}$ $Q = 4*Q_b$ .

Explanation:

Let's look at the Q on the left. It will be attracted to $Q_{b}$ $Q_b$ and repelled from the other Q. We need to find the value of Q, in terms of $Q_{b}$ $Q_b$ that will balance those 2 forces.

The equation for that state will state that the sum of those 2 forces = 0.

$\frac{k \cdot (- Q) \cdot Q_{b}}{r^{2}} + \frac{k \cdot (- Q) \cdot (- Q)}{{(2 \cdot r)}^{2}} = 0$ $(k*(-Q)*Q_b)/r^2 + (k*(-Q)*(-Q))/(2*r)^2 = 0$

$\frac{k \cdot (- Q) \cdot Q_{b}}{r^{2}} + \frac{k \cdot Q^{2}}{4 \cdot r^{2}} = 0$ $(k*(-Q)*Q_b)/r^2 + (k*Q^2)/(4*r^2) = 0$

$\frac{k \cdot Q \cdot Q_{b}}{r^{2}} = \frac{k \cdot Q^{2}}{4 \cdot r^{2}}$ $(k*Q*Q_b)/r^2 = (k*Q^2)/(4*r^2)$

$(cancel(k)*cancel(Q)*Q_b)/cancel(r^2) = (cancel(k)*Qcancel(^2))/(4*cancel(r^2))$

The result is $Q = 4*Q_b$ .

I hope this helps,
Steve

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Two identical negative charges Q are placed on equal distance r of a positive charge Qb (see photo). What is then the value of Q causing the charges (relative to each other) to be in balance? Thank you!

1 Answer

Explanation:

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