Question

four charges + Q + Q minus q and minus q and placed respectively at the corners a b c and d of a square of side a arranged in the given order calculate the electric potential at the centre o if E and F are the midpoints of sides BC and CD respectively what will be the work done in carrying a charge 'c' from O To E and from o to f​

Answer 1

Explanation:

from the figure attached below

ABCD is a square O is the center

the distance between A  to D is a

thus

BE = OE = a/2

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

then

potential is given by

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

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

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

$V_0= 0$

now , electric density is given by

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

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

thus

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

electric potential at O

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

now ,

work done is W = QV

here ,

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

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

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

thus work done

from O to f is

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

#Learn  more :

brainly.in/question/14092270

Answer 2

WOF = e(VF - Vo)

Explanation:

Eo = 2 (2EA Cos45°)

= 4 (1/4πϵo) [q/(a/√2)²] (1/√2)

= √2q/πϵoa²

Electric potential at O will be zero.  

Also VE = 0.

Word done in carrying a charge from O to E will be zero.

VF = 2 [  (1/4πϵo) [q/(a² + a²/4)^(1/2) - (1/4πϵo)(q/(a/2))]

  = (q/πϵo) [ 1/√5 - 1]

Now WOF = e(VF - Vo)