Three charges are situated as shown in the diagram (each grid line is separated

Question

Three charges are situated as shown in the diagram (each grid line is separated by 1 meter). Two of the charges (the upper left and lower right) have a charge of -3 mu C (-3 Times 10^-6 C) but the upper right charge is +4 mu C (+4 Times 10^-6 C). What is the magnitude of the net electric field at the point (0, 0)? 2,250 N/C 4,500 N/C 9,000 N/C 11,250 N/C 0 N/C What is the direction of the net electric field at the origin? up and right down and right up and left down and left zero What is the net electric potential at the point (-2, -2)? -2,250 V -4,500 V -7,136 V -19,864 V 0 V

Explanation / Answer

The both -3micro Coulomb charge are at equidistant from origin in opposite directions so the electric field produced by upper left negative charge is in opposite direcion ( with same magnitude) to the field produced by lower right negative charge .so resultant field due to these two charge will be zero since each nullified by other .so net field at origine will be only due to upper right positive charge

Hence E = kq/r2 ; k = 9×10^9 , q= 4×10^-6 C , r= 22 m

Thus E = 4500N/C

And it's direction will be away from the upper right charge .along line joining the charge to origine ( thus down and left)

Formula for potential is given by ( for a single charge at distance r)

V= kq/r

1. Due to left upper charge potential will be V1= - (9×10-9×3×10^-6/4)

2. Due to right lower charge potential will be V2 = - (9×10^-9×3×10^-6/4)

3. Due to right upper charge potential will be V3 = +(9×10×4×10^-6/ 42)

V= v1 + v2 + v3

= -7136 V

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