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Two charges 8C and 6C are placed with a distance of separation d between them and exert a force of magnitude F on each other. If a charge - 4C is added to each of these and they brought nearer by a distance d/3, the magnitude of force between them will be
- a)1/3 F
- b)9/4 F
- c)3/2 F
- d)2/3 F
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Two charges 8C and 6C are placed with a distance of separation d betwe...
Let's solve this problem step by step:
Given:
Charge 1 (q1) = 8C
Charge 2 (q2) = 6C
Distance between the charges (d) = d
Step 1: Calculating the initial force between the charges
According to Coulomb's Law, the force (F) between two charges is given by the equation:
F = (k * |q1 * q2|) / d^2
Where k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2).
Plugging in the given values, we can calculate the initial force (F_initial):
F_initial = (9 × 10^9 Nm^2/C^2 * |8C * 6C|) / d^2
= (9 × 10^9 Nm^2/C^2 * 48C^2) / d^2
= (432 × 10^9 Nm^2) / d^2
Step 2: Adding -4C charge to each charge
Now, we add a charge of -4C to each of the original charges:
q1' = q1 + (-4C) = 8C - 4C = 4C
q2' = q2 + (-4C) = 6C - 4C = 2C
Step 3: Bringing the charges closer
The distance between the charges is brought closer by a distance of d/3.
New distance = d - (d/3) = 2d/3
Step 4: Calculating the final force between the charges
Using Coulomb's Law, the final force (F_final) between the charges is given by:
F_final = (k * |q1' * q2'|) / (2d/3)^2
= (k * |4C * 2C|) / (4d^2/9)
= (k * 8C^2) / (4d^2/9)
= (9 × 10^9 Nm^2/C^2 * 8C^2) / (4d^2/9)
= (72 × 10^9 Nm^2) / (4d^2/9)
= (72 × 10^9 Nm^2 * 9) / (4d^2)
= (648 × 10^9 Nm^2) / (4d^2)
= (162 × 10^9 Nm^2) / (d^2)
Step 5: Comparing the initial and final forces
To find the magnitude of the force between them, we compare the initial and final forces:
F_final / F_initial = [(162 × 10^9 Nm^2) / (d^2)] / [(432 × 10^9 Nm^2) / d^2]
= (162 × 10^9 Nm^2) / (432 × 10^9 Nm^2)
= 0.375 = 3/2
Therefore, the magnitude of the force between the charges after the addition of -4C charge on each and bringing them closer by a distance d/3 is
Given:
Charge 1 (q1) = 8C
Charge 2 (q2) = 6C
Distance between the charges (d) = d
Step 1: Calculating the initial force between the charges
According to Coulomb's Law, the force (F) between two charges is given by the equation:
F = (k * |q1 * q2|) / d^2
Where k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2).
Plugging in the given values, we can calculate the initial force (F_initial):
F_initial = (9 × 10^9 Nm^2/C^2 * |8C * 6C|) / d^2
= (9 × 10^9 Nm^2/C^2 * 48C^2) / d^2
= (432 × 10^9 Nm^2) / d^2
Step 2: Adding -4C charge to each charge
Now, we add a charge of -4C to each of the original charges:
q1' = q1 + (-4C) = 8C - 4C = 4C
q2' = q2 + (-4C) = 6C - 4C = 2C
Step 3: Bringing the charges closer
The distance between the charges is brought closer by a distance of d/3.
New distance = d - (d/3) = 2d/3
Step 4: Calculating the final force between the charges
Using Coulomb's Law, the final force (F_final) between the charges is given by:
F_final = (k * |q1' * q2'|) / (2d/3)^2
= (k * |4C * 2C|) / (4d^2/9)
= (k * 8C^2) / (4d^2/9)
= (9 × 10^9 Nm^2/C^2 * 8C^2) / (4d^2/9)
= (72 × 10^9 Nm^2) / (4d^2/9)
= (72 × 10^9 Nm^2 * 9) / (4d^2)
= (648 × 10^9 Nm^2) / (4d^2)
= (162 × 10^9 Nm^2) / (d^2)
Step 5: Comparing the initial and final forces
To find the magnitude of the force between them, we compare the initial and final forces:
F_final / F_initial = [(162 × 10^9 Nm^2) / (d^2)] / [(432 × 10^9 Nm^2) / d^2]
= (162 × 10^9 Nm^2) / (432 × 10^9 Nm^2)
= 0.375 = 3/2
Therefore, the magnitude of the force between the charges after the addition of -4C charge on each and bringing them closer by a distance d/3 is
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Two charges 8C and 6C are placed with a distance of separation d between them and exert a force of magnitude F on each other. If a charge - 4C is added to each of these and they brought nearer by a distance d/3, the magnitude of force between them will bea)1/3 Fb)9/4 Fc)3/2 Fd)2/3 FCorrect answer is option 'C'. Can you explain this answer?
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