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Four point charges Q, 2Q,3Q and 4Q are placed at the vertices of a square of side a in order . Calculate magnitude and direction of electeic field at the point of intersection of diagonals |
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Answer» Answer: Enet= 4kQ/a^2√2 and direction is towards the side of charge Q and 2Q Explanation: question is asking to find the electric field at the center of the square. the formula for electric field is k q / SEPARATION ^2 here k is 9*10^9 separation = a/√2 electric field for Q = kQ/(a/√2)^2 = 2kQ/a^2 electric field for 2Q = 4kQ/a^2 electric field for 3Q = 6kQ/a^2 electric field for 4Q = 8kQ/a^2 now, charge Q and 3Q lies on the same diagonal and direction of electric field of QAND 3Q are opposite to each other so we can CANCEL electric field of Q nad 3Q that is remaining electric field = 4kQ/a^2 and direction is towards Q charge. similarly, we can cancel electric field of 4Q and 2Q so remaining charge = 4kQ/a^2 and direction is towards 2Q.... now you can see that there are two vector one is towards Q and other is towards 2Q and the angle between them is 90 DEGREE and both vector are of equal magnitude and that is 4kQ/a^2 ∴ electric field net = √ (4kQ/a^2)^2 + (4kQ/a^2)^2 = √2(4kQ/a^2)^2 = 4kQ/a^2√2 and direction is towards the side of Qand 2Q charge hope you understood it....................... |
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