Question

ABC is equilateral triangle of side Im. Charges are
placed at its corners as shown in fig. O is the mid-
point of side BC the potential at point (O) is-



(2) 1.52 * 10% V
(4) - 1.52 x 105 V
(1) 2.7 x 103 V
(3) 1.3 * 103 V
​

Answer 1

Answer:

send the figure to answer

Answer 2

The potential at point (O) is  -1.52x$10^{5}$ V

consider the charges:

• charge at A = -6uC
• charge at B = -3uC
• charge at C = -2uC

• distance, BO = CO = 0.5m

• distance , AO = $\sqrt{3}$/2m, AO = $\sqrt{AB^{2} - BO^{2} }$

• Total potential = Kq/r = (1/4$\pi$∈)[ -6uC/AO + -3uC/BO -2uC/CO] =

==> 9 x $10^{9}$ x { - 6uC/0.86 - 3uC/0.5 -2/0.5} = -1.52x$10^{5}$ V