Question

QUESTION 5 

two point electric charges experience a force F when they are separated by a distance d. If this distance is increased to 3d, what will the force now be? 

• F/9 

• 3F 

• F/3 

• 9F

QUESTION 6 

Two identical charges experience a repuisive force of 0.29 N when separated by a distance of 59 cm. What is the magnitude of the charges, In units of Coulombs?

Answer 1

answer 5) the electric force is given by

F=kq1q2/d²

so we have d=3d then d²=9d²

so since F and d are inversely proportional, the forces becomes 9F

so correct answer is 9F.

answer 6) we have the formula for force

F=kq1q2/r²

here q1=q2

so F=kq²/r²

0.29=8.99*10⁹*q²/0.59²

q²=11.229*10^-12

q=3.4*10^-6 C

so the answer is 3.4*10^-6 C or 3.35*10^-6C

Answer 2

SOLUTION :

5.

F is inversely proportional to  squared distance, d^2

=> F = k / d^2 

=> k = F d^2

So,

F2 = k  / (3d)^2 = F d^2 / (9 d^2) = F / 9 : First Option (ANSWER).

6.

Force between two identical charges q each :

F = k q^2 / (59/100) ^2 = 0.29 N 

=> q^2 = 0.29 *0.59^2 / k 

=> q^2 = 0.29 * 0.59^2 / (8.988 * 10^9) = 1.12315 * 10^(-11) 

 [ since, k = 8.988*10^9 n-m^2/C^2 ]

=> q = sqrt(1.12315*10^(-11)) 

=> q = 3.35 * 10^(-6) C (ANSWER).