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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?

Respuesta :

Answer:

F = 2.40 × [tex]10^{-6}[/tex]  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 × [tex]10^{-9}[/tex] × 400.72

F = 2.40 × [tex]10^{-6}[/tex]  N

onyebuchinnaji

The net force on the third charge is 2.404 x 10⁻⁶ N.

The given parameters:

  • Position of first point charge, x1 = 0.198 m
  • Position of second point charge, x2 = -0.297 m
  • First point charge, q1 = 3.95 nC
  • Second point charge, q2 = 4.96 nC
  • Third point charge, q3 = 6 nC
  • Position of the third charge, = 0

The force on the third charge due to first charge is calculated as follows;

[tex]F_{13} = \frac{kq_1 q_3}{r^2} \\\\F_{13} = \frac{9\times 10^9 \times 3.95 \times 10^{-9} \times 6 \times 10^{-9} }{(0.198)^2} (+i)= 5.44 \times 10^{-6} \ N \ (+i)[/tex]

The force on the third charge due to second charge is calculated as follows;

[tex]F_{23} = \frac{kq_2q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times 4.96 \times 10^{-9}\times 6 \times 10^{-9} }{(0.297)^2} (-i)\\\\F_{23} = (3.036 \times 10^{-6} ) \ N \ (-i)[/tex]

The net force on the third charge is calculated as follows;

[tex]F_{net} = 5.44 \times 10^{-6} - 3.036 \times 10^{-6} \\\\F_{net} = 2.404 \times 10^{-6} \ N[/tex]

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