Three unknown charges q_1, q_2, and q_3 exert forces on each other. When q_1 and

Question

Three unknown charges q_1, q_2, and q_3 exert forces on each other. When q_1 and q_2 are 15.0 cm apart (q_3 is absent), they attract each other with a force of 1.4 times 10^-2 N. When q_2 and q_3 are 20.0 cm apart (q_1 is absent), they attract with a force of 3.8 times 10^-2 N. When q_1 and q_3 are 10.0 cm apart (q_2 is absent), they repel each other with a force of 5.2 times 10^-2 N. Find the magnitude and sign of each charge. Charges q, 2q, -4q, and -2q (q is positive) occupy the four comers of a square of sides 2L, centered at the origin of .1 coordinate system (Fig. 5-1). (a) What is the net force on charge q due to the other charges? (b) What is the force on a new charge Q placed at the origin?

Explanation / Answer

4)

F = k*q1*q2/r^2

So, 1.4*10^-2 = 9*10^9*q1*q2/0.15^2

So, q1*q2 = 3.5*10^-14 C^2 ---------- (1)

Similalrly, for q2 and q3

3.8*10^-2 = 9*10^9*q2*q3/0.2^2

So, q2*q3 = 1.69*10^-13 C^2 --------- (2)

and for q3 and q1 :

5.2*10^-2 = 9*10^9*q3*q1/0.1^2

So, q3*q1 = 5.78*10^-14 C^2 ---------- (3)

Let us suppose q1 is positve, so q2 must be negative as q1 and q2 attract each other. Similarly, q3 must must be positve

Solving the three equations : we get

q1 = 1.09*10^-7 C

q2 = - 3.2*10^-7 C

q3 = 5.3*10^-7 C

5)

a)

Electric force on +q due to +2q :

F1 = k*(q)(2q)/(2L)^2 j = k*q^2/2L^2 j

Similarly, force on +q due to -2q :

F2 = -k(q)(2q)/(2L)^2 i = -k*q^2/2L^2 i

Similsrly, force on +q due to -4q :

F3 = -k*(q)(4q)/(8L^2) *(i + j)/sqrt(2) = -(k*q^2/(2*sqrt(2)L^2)) (i + j)

So, net force on +q :

Fnet = F1 + F2 + F3 = (k*q^2/2L^2) j - (k*q^2/2L^2) i - (k*q^2/(2*sqrt(2)L^2)) (i + j)

= (k*q^2/(2L^2))*[ (-1 - 1/sqrt(2)) i + (1 - 1/sqrt(2)) j ] <--------- i , j are unit vectors along +x and +y axes.

b)

Force on Q due to +q charge :

F1 = k*Qq/(2*sqrt(2)*L^2)*( -i - j)

Similalrly, force on Q due to -2q:

F2 = 2k*Qq/(2*sqrt(2)*L^2)*(-i + j)

due to -4q:

F3 = k*Q*(4q)/(2*sqrt(2)*L^2)*(- i - j)

and due to +2q:

F4 = k*Q*2q/(2*sqrt(2)*L^2)*(-i + j)

So, net force :

Fnet = (kQ*q/(2*sqrt(2)*L^2))*[ (-1 -2 -4 - 1) i + (-1 + 2 -4 + 2) j ]

Fnet = (kQ*q/(2*sqrt(2)*L^2))*[ -8 i - j ] <----- in vector notation

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