plz ans this que from physics
Answers
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☺️❤️