The drawing shows an equilateral triangle, each side of which has a length of 2.

Question

The drawing shows an equilateral triangle, each side of which has a length of 2.20 cm. Point charges are fixed to each corner, as shown. The 4.00 µC charge experiences a net force due to the charges qA and qB. This net force points vertically downward and has a magnitude of 392 N. Determine the magnitudes and algebraic signs of the charges qA and qB.

+4.00 uc
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qa qb

(sorry the picture wouldnt copy in but there are other examples of this question on cramster - I just still dont understand)

Explanation / Answer

net force head vertically down, so there is no horizontal component of force. so electric forces caused by qa and qb must have the same horizontal component's magnitude. this is an equilateral triangle so that mean qa=qb and have the same sign. -------- net force is vertically downward, so its attractive force. so that qa and qb are negative charges. --- F=k*4e-6*q/2.2e-2^2 vertical component = net force = 2*F*cos30 = 392. so that F=226(N)> from this we have q=3e-6(C)=3(uC). so qa=qb=-3uC

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