Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is at x= -0.297 m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge q3=6.00nCq that is placed at the origin?
1 answer:
Answer:
F = 2.40 × N
Explanation:
given data
charge q1 = 3.95 nC
x= 0.198 m
charge q2 = 4.96 nC
x= -0.297 m
solution
force on a point charge kept in electric field F = E × q ................1
here E is the magnitude of electric field and q is the magnitude of charge
and
first we will get here electric field at origin
So net field at origin is
E = (Kq2÷r2²) - (kq1÷r1²) ...............2
put here value
E = 9[(4.96÷0.297²)-(3.95÷0.198²)]
E = 400.72 N/C ( negative x direction )
so that force will be
F = 6 × × 400.72
F = 2.40 × N
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