Question

The electric field intensity is 400 V/m is at a distance of 2 m From apoint charge .it will be 100V/m at a distance

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf r_2=4cm}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Electric field intensity ($\sf{E_1}$)=400 V/m

• Distance ($\sf{r_1}$)=2m

• Electric field intensity ($\sf{E_2}$)=100 V/m

$\large\underline\pink{\sf To\:Find: }$

• Distance ($\sf{r_2}$)=?

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We know that :

$\large{♡}$$\large{\boxed{\sf E={\frac{1}{4πE_o}}×{\frac{q}{r^2}}}}$

$\large\implies{\sf Er^2={\frac{q}{4πE_o}} }$

$\large\implies{\sf Er^2=Constant}$

$\large{♡}$$\large{\boxed{\sf E_1r_1^2=E_2r_2^2 }}$

On Putting value :

$\large\implies{\sf 400×2^2=100×r_2^2}$

$\large\implies{\sf {\frac{400×4}{100}}=r_2^2 }$

$\large\implies{\sf r_2^2=16}$

$\large\implies{\sf r_2=4cm}$

$\huge\red{♡}$$\large\red{\boxed{\sf r_2=4cm}}$

Answer 2

Answer:

4 m

Explanation:

400 x (2)2 = 100 x r2

400 x4 /100= r2

1600 / 100=r2

16= r2

4 m = r