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Question: Electrostatic potential at H is A. \[\dfrac{\sigma }{{^{{\varepsilon _0}}}}\left[ {{{\left( {{a^2...

Electrostatic potential at H is
A. σε0[(a2H2)1/2H]\dfrac{\sigma }{{^{{\varepsilon _0}}}}\left[ {{{\left( {{a^2} - {H^2}} \right)}^{1/2}} - H} \right]
B. σε0[(a2H2)1/2+H]\dfrac{\sigma }{{^{{\varepsilon _0}}}}\left[ {{{\left( {{a^2} - {H^2}} \right)}^{1/2}} + H} \right]
C. σ2ε0[(a2+H2)1/2H]\dfrac{\sigma }{{^{2{\varepsilon _0}}}}\left[ {{{\left( {{a^2} + {H^2}} \right)}^{1/2}} - H} \right]
D. σ2ε0[(a2H2)1/2H]\dfrac{\sigma }{{^{2{\varepsilon _0}}}}\left[ {{{\left( {{a^2} - {H^2}} \right)}^{1/2}} - H} \right]

Explanation

Solution

In this question,we are going to use the concept of electrostatic potential and derive the expression of it by assuming certain assumptions.The
potential due to point charge is inversely proportional to square root of distance.

Complete step by step answer:
First let’s calculate potential due to a ring on a point lying on its axis, then we will use it to calculate potential due to disc.

Consider a small point charge on the ring having small charge dq. Now potential due to this point charge can be written as
dv=K(dq)r2dv = \dfrac{{K(dq)}}{{{r^2}}}
Total potential due to ring: V=dvV = \int {dv}
V=K(dq)r2V = \int {\dfrac{{K(dq)}}{{{r^2}}}}
r=H2+a2r = \sqrt {{H^2} + {a^2}}
V=K(dq)H2+a2V = \int {\dfrac{{K(dq)}}{{\sqrt {{H^2} + {a^2}} }}}
V=KH2+a2dqV = \dfrac{K}{{\sqrt {{H^2} + {a^2}} }}\int {dq}
V=KQH2+a2V = \dfrac{{KQ}}{{\sqrt {{H^2} + {a^2}} }}
Now, Consider a small ring in a thin disc, the small ring has radius =x = xand thickness=dx = dx. As we have already calculated the potential due to ring, we can say that potential due to the small ring is
dV=KdqH2+x2dV = \dfrac{{Kdq}}{{\sqrt {{H^2} + {x^2}} }}
Now, we can integrate thin disc to get potential of entire disc
V=dVV = \int {dV}
V=KdqH2+x2V = \int {\dfrac{{Kdq}}{{\sqrt {{H^2} + {x^2}} }}}
Now we have to substitute dq charge in small ring to thickness dx in terms of x,
dq=TotalchargeTotalarea×Areaofsmallringdq = \,\,\dfrac{{Total\,\,\,ch\arg e}}{{Total\,\,\,area}} \times Area\,\,\,of\,small\,\,ring
dq=Qπa2×(2πx)×dxdq = \dfrac{Q}{{\pi {a^2}}} \times \left( {2\pi x} \right) \times dx
Area of ring is calculated by approximating small ring as rectangle of length equal to circumference of small ring and height equal to the thickness
V=KQ2πx(πa2)(H2+x2)dxV = \int {\dfrac{{KQ\,\,\,2\pi x}}{{(\pi {a^2})\,(\sqrt {{H^2} + {x^2}} )}}} dx
V=KQa22xdxH2+x2V = \dfrac{{KQ}}{{{a^2}}}\int {\dfrac{{2xdx}}{{\sqrt {{H^2} + {x^2}} }}}
Here the limit of xx is from 00 to aa
So, V=KQa20a2xdxH2+x2V = \dfrac{{KQ}}{{{a^2}}}\int_0^a {\dfrac{{2xdx}}{{\sqrt {{H^2} + {x^2}} }}} ….. (i)
Now, SubstitutingH2+x2=t{H^2} + {x^2} = t, putting for lower limit x=0x = 0then H2+02=t{H^2} + {0^2} = t t=H2 \Rightarrow t = {H^2} now for upper limit x=ax = a then H2+a2=t{H^2} + {a^2} = t,
H2+x2=t{H^2} + {x^2} = t
Differentiating, tt with respect to xx so, 2x dx=dt2x{\text{ }}dx = dt
Putting all above values in equation (i).
V=KQa2H2H2+a2dttV = \dfrac{{KQ}}{{{a^2}}}\int_{{H^2}}^{{H^2} + {a^2}} {\dfrac{{dt}}{{\sqrt t }}}
V=KQa22(H2+a2H2) V=KQa22(H2+a2H)  V = \dfrac{{KQ}}{{{a^2}}}2\left( {\sqrt {{H^2} + {a^2}} - \sqrt {{H^2}} } \right) \\\ V = \dfrac{{KQ}}{{{a^2}}}2\left( {\sqrt {{H^2} + {a^2}} - H} \right) \\\
Putting K=14πε0K = \dfrac{1}{{4\pi {\varepsilon _0}}} and σ=Qπa2\sigma = \dfrac{Q}{{\pi {a^2}}}, we get
V=σ2εo(H2+a2H)V = \dfrac{\sigma }{{2\varepsilon o}}\left( {\sqrt {{H^2} + {a^2}} - H} \right)
Here option (C) is the correct option.

Note: As the limits of xxis from 00to aaand H2+x2=t{H^2} + {x^2} = t
at x=0, t=H2x = 0,{\text{ t}} = {H^2}
at x=a, t=H2+a2x = a,{\text{ t}} = {{\text{H}}^2} + {a^2}
So, the limits of ttwill be from H2{H^2}to H2+a2{H^2} + {a^2}