Two equal charges are placed in air 10cm apart and the force of intraction between them is 50mg-wt. Determine the magnitude of the charges?

please solve this
send me the solution?

Question

Two equal charges are placed in air 10cm apart and the force of intraction between them is 50mg-wt. Determine the magnitude of the charges?please solve thissend me the solution?

Answer

Solution:

Given:
Distance between the two equal charges, d = 10 cm = 0.1 m
Force of interaction between the two equal charges, F = 50 mg-wt = 50 × 10^-3 × 9.8 N

To find: Magnitude of the charges, q

Coulomb’s law states that the force of interaction between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Therefore, the formula for Coulomb’s law is given by:

F = k q1 q2 / r^2

where k is the Coulomb’s constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

Substituting the given values in the above equation, we get:

50 × 10^-3 × 9.8 = k q^2 / (0.1)^2

Solving for q, we get:

q = √[(50 × 10^-3 × 9.8 × 0.1^2) / k]

where k = 9 × 10^9 N m^2 C^-2 is the Coulomb’s constant.

q = √[(50 × 10^-3 × 9.8 × 0.1^2) / 9 × 10^9]

q = √[0.00000544]

q = 0.00233 C

Therefore, the magnitude of each of the two equal charges is 0.00233 C.

Answer

Unit of given force is wrong it is newton(kg-m/s^2). to solve this use formula : F=(k*q*q')/r^2..
k=9*10^9N-m^2/c^2
q=q' given
put all value and take care of unit and solve it.

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