Question

electric intensity at a point 50 cm of formal charge of 3.2 microcoulomb in a medium of dielectric constant 2 is equal to​

Answer 1

Answer:

electric intensity at a point 50 cm of formal charge of 3.2 microcoulomb in a medium of dielectric constant 2 is equal to​ 57600 N/C

Explanation:

As we know that the electrical field intensity due to a point charge is given as

$E = \frac{q}{4\pi \epsilon_r \epsilon_0 r^2}$

here we know that

$\epsilon_r = 2$

$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$

now electric field is given as

$E = \frac{(9\times 10^9)(3.2 \times 10^{-6})}{2(0.50)^2}$

$E = 57600 N/C$

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Topic : Electric field due to a point charge

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Answer 2

Answer:

The electric intensity at a point 50 cm of formal charge of 3.2 microcoulomb in a medium of dielectric constant 2 is  ​$=57.6*10^{3} N/C$

Explanation:

The electric field is the electric force per unit charge

The electric field due to a charge $q$ at a distance $r$ from the charge in a medium of dielectric constant  $k$ is given by the formula

$E=\frac{1}{4\pi \varepsilon _0k}\frac{q}{r^2}$

the value of $\frac{1}{4\pi \varepsilon _0}=9\times 10^9$

From the question, the given values are

charge $q=3.2^{}\mu C =3.2*10^{-6}C$

distance $r=50cm=0.50m$

the dielectric constant of the medium $k=2$

substitute the given values to get the electric field

⇒$E=\frac{9*10^{9}*3.2*10^{-6} }{2*0.5^{2} }$

      $=57.6*10^{3} N/C$