Four point charges of equal magnitude are held at the corners of a square as shown below. The length of each side of the square is 2a and the sign of the charges is as shown. The point P is at the centre of the square. a) On the diagram above, draw an arrow to represent the direction of the resultant electric field at point P. b) Show that the magnitude of the electric field strength at point P due to one of the point charges is equal to (kQ/2a^2). c) Determine, in terms of k, Q and a, the magnitude of the total electric field strength at point P.

Four point charges of equal magnitude are held at the corners of a square as shown below. The length of each side of the square is 2a and the sign of the charges is as shown. The point P is at the centre of the square. a) On the diagram above, draw an arrow to represent the direction of the resultant electric field at point P. b) Show that the magnitude of the electric field strength at point P due to one of the point charges is equal to (kQ/2a^2). c) Determine, in terms of k, Q and a, the magnitude of the total electric field strength at point P.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter2: Motion In One Dimension

Section: Chapter Questions

Problem 2.15OQ: A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 4...

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