Physics, asked by krish0258, 2 months ago

plz ans this que from physics​

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Answers

Answered by prskoranga
1

Answer:

ABCD is a square O is the center

the distance between A to D is a

thus

BE = OE = a/2

and BO = \frac{a}{\sqrt{2}}

2

a

then

potential is given by

V = \frac{KQ}{R}V=

R

KQ

where k is the constant Q is the charge and R is the distance

V_o = \frac{KQ}{\frac{a}{\sqrt{2}} }+\frac{KQ}{\frac{a}{\sqrt{2}} }-\frac{KQ}{\frac{a}{\sqrt{2}} }-\frac{KQ}{\frac{a}{\sqrt{2}} }V

o

=

2

a

KQ

+

2

a

KQ

2

a

KQ

2

a

KQ

V_0= 0V

0

=0

now , electric density is given by

E = \frac{KQ}{R^2} }

\frac{KQ}{(\frac{a}{\sqrt{2})} )^2}= \frac{2KQ}{a^2}

(

2

)

a

)

2

KQ

=

a

2

2KQ

thus

E = \sqrt{(\frac{4KQ}{a^2})^2+(\frac{4KQ}{a^2})^2}E=

(

a

2

4KQ

)

2

+(

a

2

4KQ

)

2

electric potential at O

E = \frac{4\sqrt{2} KQ}{^2}E=

2

4

2

KQ

now ,

work done is W = QV

here ,

W from (O to E) = W from (O to J)

V_f =\frac{KQ}{\frac{\sqrt{5} }{2}a } +\frac{KQ}{\frac{\sqrt{5} }{2}a } -\frac{KQ}{\frac{a}{2} }-\frac{KQ}{\frac{a}{2} }V

f

=

2

5

a

KQ

+

2

5

a

KQ

2

a

KQ

2

a

KQ

V_f = \frac{4KQ}{a} (\frac{1}{\sqrt{5} }-1)V

f

=

a

4KQ

(

5

1

−1)

thus work done

from O to f is

\begin{gathered}W = +Q(V_f-v_o)\\\\W = Q\times \frac{KQ}{a}(\frac{4}{\sqrt{5} }-4) \\\\W = \frac{4KQ}{a} (\frac{1}{\sqrt{5} }-1)\end{gathered}

W=+Q(V

f

−v

o

)

W=Q×

a

KQ

(

5

4

−4)

W=

a

4KQ

(

5

1

−1)

Hope it helps☺️❤️

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