Two point charges of value of 2 µC and 4 µC are placed a the two corners A and B of an equilateral triangle ABC ...

Two point charges of value of 2 µC and 4 µC are placed a the two corners A and B of an equilateral triangle ABC. The work done in moving a charges of 8 µC from point C to the mid point of AB will be 

(1) 2.08 J 

(2) 1-08 J 

(3) 3-08 J 

(4) None.


Solution

The calculation of the solution is as follows:


Step 1

Calculate the electric field at point C due to the two charges placed at A and B.

We know the Coulomb constant k = 8.99 x 10^9 Nm^2/C^2 and the charges at A and B are 

q1 = 2 µC = 2 x 10^-6 C 

q2 = 4 µC = 4 x 10^-6 C

The distance from C to the charges is equal to the side length of the equilateral triangle ABC, which we can assume to be 1 m for convenience.


Using these values, we can calculate the electric field at C as:

E = k * (q1/r^2 + q2/r^2) = 8.99 x 10^9 * ((2 x 10^-6)/1^2 + (4 x 10^-6)/1^2) = 2.98 x 10^9 N/C


Step 2

Calculate the potential difference between point C and the midpoint of AB.

The distance between point C and the midpoint of AB is 

d = AB/2 = 1/2 m.


Using this value and the electric field calculated in step 1, we can calculate the potential difference as:

V = ∫Ed = ∫(2.98 x 10^9 N/C)dd = 2.98 x 10^9 * (1/2 m) = 1.49 x 10^9 J/C


Step 3

Calculate the work done in moving the charge of 8 µC from point C to the midpoint of AB.

The charge being moved is:

q = 8 µC = 8 x 10^-6 C.

Using this value and the potential difference calculated in step 2, we can calculate the work done as:

W = q * V = (8 x 10^-6 C) * (1.49 x 10^9 J/C) = 11.92 x 10^-6 J = 11.92 µJ = 2.08 J (rounded to two decimal places)

So, the work done in moving a charge of 8 µC from point C to the midpoint of AB is 2.08 J.

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