Equal charges each of 20 meucoulumb …
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Payal Singh 6 years, 10 months ago
Solution:
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 = {tex}k {q_1q_2\over (r_{21})^2} + k {q_3q_2\over (r_{23})^2}+ k {q_1q_4\over (r_{24})^2}+ k {q_1q_5\over (r_{25})^2}{/tex}
{tex}F = k {qq\over (r_{21})^2} + k {qq\over (r_{23})^2}+ k {qq\over (r_{24})^2}+ k {qq\over (r_{25})^2}{/tex}
[q1=q2 =q3 =q4 = q5 = q = {tex}20\times 10^{-6} C{/tex}]
{tex}F = k {q^2\over (r_{21})^2} + k {q^2\over (r_{23})^2}+ k {q^2\over (r_{24})^2}+ k {q^2\over (r_{25})^2}{/tex}
{tex}F = k q^2\left[{1\over (r_{21})^2} + {1\over (r_{23})^2}+ {1\over (r_{24})^2}+ {1\over (r_{25})^2}\right]{/tex}
{tex}r_{21} = 2cm = 2\times 10 ^{-3}m\\ r_{23} = 2cm = 2\times 10 ^{-3}m \\ r_{24} = 6cm = 6\times 10 ^{-3}m\\ r_{25} = 14cm = 14\times 10 ^{-3}m{/tex}
Putting values of all, we get
{tex}F = 9\times10^9 \times 20\times 10^{-6}\left[{1\over (2\times 10^{-2})^2} + {1\over (2\times 10^{-2})^2}+ {1\over (6\times 10^{-2})^2}+ {1\over (14\times 10^{-2})^2}\right]{/tex}
F = {tex}1.92\times 10^4 N{/tex}
4Thank You