Question

Three point charges are located in free space as follows: QI
6C at Pl (1.0.0), Q2 = 10 C at P2 (2,0,0) & Q3 - 4 C at
P3 (4,0,0)
Which charge has greatest magnitude of force on it? What is the
magnitude of that force?​

Answer 1

Answer:

Explanation:

Q

2

​

and Q

3

​

are equal is magnitude but opposite in sign.

∴Q

2

​

and Q

3

​

experience attractive force due to each other. Now Q

1

​

must have sign of Q

3

​

to repel it so that Net force on Q

3

​

is zero

⇒

r

2

−kQ

2

​

Q

3

​

​

=

(2r)

2

k(Q

1

​

)Q

3

​

​

⇒Q

1

​

=−4Q

2

​

⇒Q

1

​

=4∣Q

3

​

∣

as ∣Q

2

​

∣=∣Q

3

​

∣

Answer 2

Answer:

10C charge has greatest magnitude of force on it and its magnitude is

63 \times 10 {}^{10} N[/tex]

Explanation:

Given that,The 3 point charges are as follows

q_{1}=6C \\ q_{2}=10C \\ q_{3}=-4C

And these point charges are located in free space at the given point as follows

(1,0,0),(2,0,0) and (4,0,0)

By using the electrostatic formular given below we can calculate the forces

$F=k_{e}\frac{q_{1}q_{2}}{r^{}{2}}$

The force are calculated as below

$F_{1}=(9 \times 10 {}^{9}) \times \frac{q_{1}q_{2}}{r_{12}^{2}} \\ = 9 \times 10 {}^{9} \times \frac{6 \times 10}{1 {}^{2} } = 54 \times 10 {}^{10} N$

F_{2}=(9 \times 10 {}^{9}) \frac{q_{1} |q_{3} | }{r_{13}^{2}} \\ = 9 \times 10 {}^{9} \times \frac{6 \times 4}{3 {}^{2} } = 2.4 \times 10 {}^{10} N \\

F_{ 3}=(9 \times 10 {}^{9}) \frac{q_{2} |q_{3} | }{r_{23}^{2}} \\ = 9 \times 10 {}^{9} \times \frac{10 \times 4}{2 {}^{2} } = 9 \times 10 {}^{10} N

The net force on 6C Charge is

$F_{6C}=F_{1}-F_{2} \\ = (54 - 2.4) \times 10 {}^{10} = 51.6 \times 10 {}^{10} N$

Similarly for other charges,

$F_{10C}=F_{1} + F_{3} \\ = (54 + 9) \times 10 {}^{10} = 63 \times 10 {}^{10 } N$

$F_{ - 4C}=F_{2} + F_{3} \\ = (2.4 + 9) \times 10 {}^{10} = 11.4 \times 10 {}^{10} N$

Therefore,10C charge has greatest magnitude of force on it and its magnitude is

63 \times 10 {}^{10} N[/tex]

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