Question

Question 11 (1 point) There is a 4nC located at the position (2m.0) and a -6nC charge located at (0,3m). What is the electric potential V at the origin? (9210°) (4:10) (9–10°) (6x10-9) V = 18 + 18 = 36J/C + 2 3 You can not add them because one is from the x-axis and the other is from the y- axis. ov (9x10') (4x10") 22 (9:10")( 6x109) = 9 - 6 = 3J/C V + (9x10°) (4x10') (9x10°) (-6x109) V + = 18 - 18 = 0J/C 3 2 Page 11 of 18

can you just answer the forst one then

Answer 1

Solution :

Here we have :

Q+ = 4 nC = 4 x 10^-9 C

Q- = - 6 nC = - 6 x 10^-9 C

(-6 nc 3 m 2 m 4 nc Origin

Here, Electric potential at the origin will be given by the sum of the potential due to each charges at the origin.

$\therefore V=\frac{kQ_+}{x}+\frac{kQ_-}{y}$

$\therefore V=\frac{(9\times 10^{9})(4\times 10^{-9}\ C)}{(2\ m)}+\frac{(9\times 10^{9})(-6\times 10^{-9}\ C)}{(3\ m)}$

$\mathbf{\therefore V=18\ V-18\ V=0}$

.

Therefore, Answer will be :

$\bullet \ V=\frac{(9\times 10^{9})(4\times 10^{-9})}{(2)}+\frac{(9\times 10^{9})(-6\times 10^{-9})}{(3)}=18-18=0\ J/C$