Question

A point charge produces an electric field of magnitude 2N/C at a point distant 0.25m from it what is the
value of charge?
1) 1.39 x 10-110
2) 1.39 x 10110
3) 13.9 x 10-110
4) 13.9 x 10110
0 1
02
O4​

Answer 1

Answer:

• Electric field (E) = 2 N/C
• Distance (r) = 0.25
• charge (Q) = ?

Electric Field due to point charge :

$\longrightarrow\sf E = \dfrac{KQ}{r^2}$

$\longrightarrow\sf E = \dfrac{1}{4\pi \varepsilon_{0}} . \dfrac{Q}{r^2}$

$\longrightarrow\sf 2=9 \times {10}^{9} . \dfrac{Q}{(0.25)^2}$

$\longrightarrow\sf 2=9 \times {10}^{9} . \dfrac{Q}{0.0625}$

$\longrightarrow\sf 2 \times 0.0625=9 \times {10}^{9} \times Q$

$\longrightarrow\sf 0.125=9 \times {10}^{9} \times Q$

$\longrightarrow\sf 125 \times {10}^{ - 3} =9 \times {10}^{9} \times Q$

$\longrightarrow\sf Q = \dfrac{125 \times {10 }^{ - 3} }{9 \times {10}^{9} }$

$\longrightarrow\sf Q = \dfrac{125 \times {10 }^{ - 3} \times {10}^{ - 9} }{9 }$

$\longrightarrow\sf Q = \dfrac{125 \times {10 }^{ - 12} }{9 }$

$\longrightarrow \underline{ \boxed{\bold{ Q = 13.9 \times {10 }^{ - 12} \: C}}}$