11. Four equal charges 2PC each are placed at x-axis just at
positions respectively x = 0,2,4 and 8 cm find out force
on charge at x = 2 cm in newton -
(1) 0
(2) 5
(3) 10
(4) 10-2
Answers
Answered by
1
Answer:
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.
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=2×10−3mr23=2cm=2×10−3mr24=6cm=6×10−3mr25=14cm=14×10−3m
Putting values of all, we get
F=9×109×20×10−6[1(2×10−2)2+1(2×10−2)2+1(6×10−2)2+1(14×10−2)2]
F = 1.92×104N
Answered by
7
Answer:
꧁༒ÂⱠÊӾ ᴳᵒᵈ༒꧂
⌛⏳FOLLOW Me ✔️⚡
1.92...❤
#TRUE BRAINLIAN#BE BRAINLY♻️
❌✖️FREE FORM SPAM✖️❌
✔️MARK MY ANSWERS BRAINLIEST
Similar questions