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professor190 [17]
4 years ago
14

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

the charges Q3=2Q and Q4=7Q that are a distance d/3 apart?
Physics
1 answer:
gladu [14]4 years ago
7 0

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

F = \frac{KQ_{1}Q_{2}}{d^{2}}

1.60 = \frac{4KQ^{2}}{d^{2}}    .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

F' = \frac{9KQ_{3}Q_{4}}{d^{2}}

F' = \frac{126KQ^{2}}{d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

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Read 2 more answers
What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field
earnstyle [38]

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

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E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

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Therefore, the electric field is  3.91 x 10⁶ N/C

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3 years ago
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