Question

q1 =3.1×10^-19c , q2 =2×10^-18c,r=2.5m , f=?​

Answer 1

AnsWer :

$\bullet \: \tt Q_1 = 3.1 \times 10^{-19} \: C$

$\bullet \: \tt Q_2 = 2 \times 10^{-18} \: C$

$\bullet \: \tt R = 2.5\: m$

$\maltese\: \underline{\textbf{According to the Question :}}$

$\longrightarrow\:\sf F =K \dfrac{Q_1Q_2}{R^2}$

$\longrightarrow\:\sf F = \dfrac{1}{4\pi \varepsilon _{0} } \cdot\dfrac{Q_1Q_2}{R^2}$

$\longrightarrow\:\sf F = 9 \times {10}^{9} \cdot\dfrac{3.1 \times {10}^{ - 19} \times 2 \times {10}^{ - 18} }{(2.5)^2}$

$\longrightarrow\:\sf F = 9 \times {10}^{9} \cdot\dfrac{6.2\times{10}^{ - 37} }{6.25}$

$\longrightarrow\:\sf F =\dfrac{55.8\times{10}^{ - 28} }{625 \times {10}^{ - 2} }$

$\longrightarrow\:\sf F =\dfrac{558 \times {10}^{-1}\times{10}^{ - 28} \times {10}^{ 2}}{625 }$

$\longrightarrow\:\sf F =\dfrac{558\times{10}^{ - 27} }{625 }$

$\longrightarrow\:\sf F =0.8928\times{10}^{ - 27}$

$\longrightarrow\:\sf F =8928\times {10}^{ - 4} \times{10}^{ - 27}$

$\longrightarrow\: \underline{ \underline{\sf F =8928\times {10}^{ - 31} \: N}}$