Equal charges each of 20 microcoulomb are placed at x=0,2,4,8,16cm on x-axis. Find the force experienced by the charge at x=2cm
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Answered by
41
Answer:
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= 1.92×10-4N
Answered by
8
Answer:
F=[Q^2*K] [1/(0.08-0.02)^2+1/(0.16-0.02)^2=1.2*10^3 ANSWER
Step-by-step explanation:
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