Question

Equal charges of each 2micro coulomb are placed at a point x=0,2,4 and 8 cm on the X-axis. the force experienced by the charge at x=2 cm is equal to:

(A)5N

(B)10N

(C)0N

(D)15N​

Answer 1

Step-by-step explanation:

Let q1,q2 ,q3 ,q4 and q5  be the charges at x = 0,2,4,8,16 on x-axis.

Let f be the resultant force.  Thus,

f1 - Force on q2 due to q1

f3 - Force on q2 due to q3

f4 - Force on q2 due to q4

f5 - Force on q2 due to q5

f = f1+f3+ f4+ f5

f = kq1q2/(r21)²+kq3q2/(r23)²+kq1q4/(r24)²+kq1q5/(r25)²

f=kqq/(r21)²+kqq/(r23)²+kqq/(r24)²+kqq/(r25)²

= q1=q2 =q3 =q4 = q5  = q = 20×10−6C

f =kq²/(r21)²+kq²/(r23)²+kq²/(r24)²+kq²/(r25)²

f=kq²[1/(r21)²+1/(r23)²+1/(r24)²+1/(r25)²]

r21=2cm=2×10−3m

r23=2cm=2×10−3m

r24=6cm=6×10−3m

r25=14cm=14×10−3m

Substituting the values -

f=9×10-9×20×10−6 [1(2×10−2)²+1(2×10−2)²+1(6×10−2)²+1(14×10−2)²]

f= 1.92×10-4 N