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jolli1 [7]
3 years ago
10

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?
Physics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

F = 2.40 × 10^{-6}  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 × 10^{-9} × 400.72

F = 2.40 × 10^{-6}  N

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