Question

of
original value and
charge
If the source charge
source charge is becomes
twice
distance between
source
becomes half of original
value then what is a new Ef
Consider initial Ef = {
& test charge​

Answer 1

Answer:

As E=4πε01r2q

If q′=2q, then E′=4πε01r22q or E′=2E

So electric field is doubled

Answer 2

Appropriate Question;

If the source charge is becomes twice of original value and distance between source charge and test charge becomes half of the original value, then what is new electric field. {Consider the initial electric field as E}.

Solution;

Electric field due to a point charge:

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

Electric field is directly proportional to the source charge and inversely proportional to the square of separation between them.

$\longrightarrow\:\sf E \propto\dfrac{Q}{r^2}$

According to the Question;

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{ \dfrac{ Q}{ {r}^{2} } }{ \dfrac{2Q}{ \bigg( \dfrac{ r }{2} \bigg)^{2} } }$

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{ \dfrac{ Q}{ {r}^{2} } }{ \dfrac{2Q}{ \dfrac{ {r}^{2} }{4} } }$

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{ \dfrac{ Q}{ {r}^{2} } }{ \dfrac{2Q \times 4}{ {r}^{2} }}$

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{ \dfrac{ Q}{ {r}^{2} } }{ \dfrac{8Q }{ {r}^{2} }}$

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{ Q}{ 8Q }$

$\longrightarrow\:\sf \dfrac{E}{E'} = \dfrac{1}{ 8 }$

$\longrightarrow\: \underline{ \boxed{ \orange{\bold{ E' = 8E}}}}$