Question

Would anyone mind explaining how to do this problem?

Answer 1

$T\cos \theta =mg$

$T\sin \theta =|q|E$

$\tan \theta =\frac {qE}{mg}$

$\tan 10^{\degree} =\frac {|q|\cdot 4000 }{0.60\cdot 10^{-3}\cdot 9.8}$

$|q|=2.59\times 10^{-7}\: C$ (since the ball goes opposite to the applied electric field, it must have a negative charge.)

$q=-2.59\times 10^{-7}\: C$