Question

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has components 1.2 x 10^-3 N directed north, 5.7 x 10^-4 N directed east and 2.2 x 10^-4 N directed vertically upward. The charge on this ball is 110 nC.
If this ball were replaced with one that has a charge of -50nC , what would be the force components exerted one the replacement ball?

F north=?
F east=?
F up=?​

Answer 1

We have that the values for F north, F east, F up are

• $F_N=1.09090909*10^{-5}$
• $F_E=5.18181818*10^{-6}$
• $F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north  is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

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