Two small spheres are given identical positive charges. When they are 1 cm apart, the repulsive force on each?

Question

of them is 0.002 N. What would the force be if (a) the distance is increased to 3 cm? (b) one charge is doubled? (c) both charges are tripled? (d) one charge is doubled and the distance is increased to 2 cm?

If you could just help me figure out which formula to use, I could solve the problem.

Answer 1

Coulombs law
The force is inverse proportional to the square of their distance d.
If d 3x bigger, then F 3^2 times smaller
The force is proportional to each charge.

Answer 2

Use Coulomb’s law .

F = (8.99 x10^9) Q1 X Q2 / R^2.

a). Since, charges remain the same, force is inversely proportional to the square of the distance between them.

F2/F1 = (R1/R2)^2. (R1/R2)^2 = (1/3)^2 = 1/9.

F2 = 0.002 x (1/9) =0.00022N.

b). One charge is doubled. Other things remain the same and hence force is doubled. The force is 2 x 0.002N.

c). Both charges are tripled. Other things remain the same and hence force is multiplied by 3x3(= 9). The force is 9 xz 0.002 N.

d). One charge is doubled. The force is doubled. The force is 2F. The distance is increased 2 times (2 cm). The force is ¼ x 2F = 0.5 F = 0.5 x 0.002 N.

Answer 3

the force between any two charges is=(q1*q2)/d*d...........i.e.,the product of the charges (vatever may be the charges positive or negative or positive-negative)where q1-charge at one end & q2-charge at the other end divided by the suare of the distance(d) between them..........& in some cases this total value is multiplied by the permitivity (e-the no.of charges that the medium can permit).if the charges are tripled then multiply the charges & divide it by the square of the total distance between them.since the force is given in this case just form a general equation of the forces for the basic question then form another equation for (a),(b),(c) seperately. remember the charges,the medium are all same for the whole question so solve out the euations one by one & find the answer
(A)force is .002N
so the general equn is F=(q1*q2)/d*d=q1*q2(since d=1,d*d=1)
now F=q1*q2---------(1) assume this as equn (1)
(leave the value e since it is same for all the three it cancels out while solving)
now given the distance is increased by 3cm
now, F=(q1*q2)/3*3=q1*q2/9
q1*q2=9F--------(2)
equating (1)&(2)
(q1*q2)/.002=(q1*q2)/9F
therefore F=.002/9
(NOW FIND FOR THE REMAINING IN THE SAME WAY)
ALL THE BEST

Answer 4

I believe the force varies with the cube of the distance and linearly with charge.

Answer 5

the formula for this is F= (k*a*b)/(d^2) where k is constant(9E9) and a and b are the measure of charge in columbs and d is the distance between the center od a and b.

Answer 6

use coulombs law.take the spheres as point charge as they r very "small n identical"

Answer 7

Thats a good question man....