Answer:
5.92 *10^-6 C
Explanation:
For the two charges q3 the force between them is given by
[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]
Now we know that
F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives
[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]
Now we solve for q_3.
Dividing both sides by 8.99 * 10^9 gives
[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]
multiplying both sides by (0.0809)^2 gives
[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]
finally, taking the square root of both sides gives
[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]
Evaluating the right-hand side gives
[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]
Hence, the charge q_3 is 5.92 x 10^-6 C.
