Electric field intensity at a point at a distance 60 cm from the charge is to voltmeter the charge will be
Answers
Answer:
charge is given by:
\boxed{E = \frac{1}{4\pi\varepsilon_{\circ}}\frac{q}{r^2}}
E=
4πε
∘
1
r
2
q
Where
qq = Charge
rr = Distance from point charge.
We are given that the Electric Field Intensity is 400 V/m at a distance of 2 m. We have to find at what distance the Electric Field will be 100 V/m.
Let that distance be rr .
Here we are not altering the charge. So we can say:
\begin{lgathered}E = \frac{1}{4\pi\varepsilon_{\circ}} \frac{q}{r^2} \\ \\ \\ \implies E \, r^2 = \frac{q}{4\pi\varepsilon_{\circ}} \\ \\ \\ \implies E \, r^2 = constant \\ \\ \\ \implies E_1 r_1 ^2 = E_2 r_2 ^2 \\ \\ \\ \implies 400 \times (2)^2 = 100 \times r^2 \\ \\ \\ \implies r^2 = \frac{400 \times 4}{100} \\ \\ \\ \implies r^2=16 \\ \\ \\ \implies \boxed{r=4 \, \, m}\end{lgathered}
E=
4πε
∘
1
r
2
q
⟹Er
2
=
4πε
∘
q
⟹Er
2
=constant
⟹E
1
r
1
2
=E
2
r
2
2
⟹400×(2)
2
=100×r
2
⟹r
2
=
100
400×4
⟹r
2
=16
⟹
r=4m
Thus, The Electric Field Intensity will be 100 V/m at a distance of 4 metres from the point charge.
So, the answer is Option (c) 4 m.