Question

Electric field strength due to a point charge of 5uC at a
distance 80 cm from the charge is
(a) 8x10" NC -1
(b) 7 x 104 NC-1
(c), 5 x 10" NC-1
(d) 4x10° NC-1​

Answer 1

AnswEr :

Option (B) is correct

From the Question,

• Charge (Q) = 5uC

• Distance of Separation (r) = 80 cm.

We have to find the electric field due to the point charge

Electric Field due to a point charge is given as :

$\large{ \star \: \boxed{ \boxed{ \sf E = \dfrac{KQ}{r^2} }}}$

Also,

• 1 uC = 10^{-6} C

• 1 cm = 10^{-1} m

Thus,

$\sf \: E = \dfrac{9 \times {10}^{9} \times 5 \times {10}^{ - 6} }{(8 \times {10}^{ - 1} ) {}^{2} } \\ \\ \longrightarrow \: \sf \: E = \dfrac{45 \times {10}^{3} }{64 \times {10}^{ - 2} } \\ \\ \longrightarrow \: \sf \: E = 0.703 \times {10}^{5} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf E = 7 \times {10}^{4} \: NC {}^{ - 1} }}$

The electric field due to the point charge is 7 × 10⁴ N/C

Answer 2

Answer:

• Electric field strength (E) is 7.04 × 10⁴ N/C

Given:

1. Charge given (Q) = 5 μC
2. Distance (d) = 80 cm = 0.8 m

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\bigstar\;\underline{\boxed{\sf E=\dfrac{1}{4\;\pi\;\epsilon_{0}}\;.\;\dfrac{Q}{r^{2}}}}$

Here,

• E Denotes Electric field intensity
• Q Denotes charge
• r Denotes Distance

Substituting the values,

$\displaystyle\longrightarrow\sf E=\dfrac{1}{4\;\pi\;\epsilon_{0}}\;.\;\dfrac{5\;\mu C}{(0.8)^{2}}$

• $\Bigg\lgroup \bf 1\;\mu\;C=10^{-6}\;C\Bigg\rgroup$

$\displaystyle\longrightarrow\sf E=\dfrac{1}{4\;\pi\;\epsilon_{o}}\;.\;\dfrac{5\times 10^{-6}}{(0.8)^{2}}\\\\\\\longrightarrow\sf E=\dfrac{1}{4\;\pi\;\epsilon_{o}}\times \dfrac{5\times 10^{-6}}{0.64}\\\\\\\longrightarrow\sf E=\dfrac{1}{4\;\pi\;\epsilon_{o}}\times \dfrac{5\times 10^{-6}\times 10^{2}}{64}\\\\\\\longrightarrow\sf E=\dfrac{1}{4\;\pi\;\epsilon_{o}}\times \dfrac{5\times 10^{-4}}{64}$

• $\Bigg\lgroup\bf \dfrac{1}{4\;\pi\;\epsilon_{o}}=9\times 10^{9}\Bigg\rgroup$

Substituting and solving,

$\displaystyle\longrightarrow\sf E=9\times 10^{9}\times \dfrac{5\times 10^{-4}}{64}\\\\\\\longrightarrow\sf E=45\times 10^{\;(9-4)}\times \dfrac{1}{64}\\\\\\\longrightarrow\sf E=\dfrac{45\times 10^{5}}{64}\\\\\\\longrightarrow\sf E=0.703\times 10^{5}\\\\\\\longrightarrow\sf E=7.03 \times 10^{4}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf E=7.03\times 10^{4} \; N/C}}}}$

∴ Electric field strength (E) is 7.03 × 10⁴ N/C.

Hence Option-b is correct !

$\rule{300}{1.5}$