find the electric field of point 'O' which is the center of a square of side 2cm ...the charges are placed on the corners of the square having a magnitude +2Q , -2Q , +Q , -Q ... Q = 0.02 * 10^-6 C ... K = 9*10^9 NM^2C^-2
Asked by Zubin Khalfay | 1st May, 2013, 12:46: AM
Lets ABCD be a square where the charges +2Q, -2Q, +Q and -Q are placed at A,B,C and D respectively. Let O be the centre of the square where we have to find the electric field. Let E be the mid point of side BC.
Now For the point O, A and C are at an equal distance on either side. Charges +2Q is placed at A and +Q is at B. So due to these 2 charges, the electric field will be along OC and the effect will be that of 2Q-Q = Q.
Similarly for the negative charges -2Q and -Q are placed at B and D. Due to these charges, the electric field will be -Q along OB.
So the total electric field will be the addition of fields along OB and OC, which comes out to be in the direction of OE.
Net electric field will be Ecos45+Ecos45 =
2Ecos45 = ?2 E = ?2KQ/OA2 = ?2 x 9x109x 0.02x10-6 / (?2)2
= 127.27 N/C along OE.
Answered by | 8th May, 2013, 09:24: PM
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