00:01
Hi everyone, here it is given three point charges, q1, 1 into 10 to the power 6 column, q2, 2 into 10 to the power 6 column, q3, 3 into 10 to the power 6 column.
00:20
At the vertices of triangle, we have to find the potential at p.
00:30
Potential at p due to q1, q2, q3.
00:34
We have to apply the principle of superposition, kq1 by r1.
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Kq2 by r2, k q3 by r3.
00:46
Substitute the value.
00:48
K will be common, 9 into 10 to the power 9...