English
Hindi
2. Electric Potential and Capacitance
hard

બે સમકેન્દ્રિય ગોળીય કવચથી કેપેસિટર બનાવવામાં આવે છે, ${R_1}$ ત્રિજયાવાળી ગોળીય કવચનો વોલ્ટેજ ${V_1}$ અને ${R_2}$ ત્રિજયાવાળી ગોળીય કવચનો વોલ્ટેજ ${V_2}$ છે,તો કેન્દ્રથી $x$ અંતરે આવેલા બિંદુએ વોલ્ટેજ કેટલો થાય? (${R_2} > x > {R_1}$)

A

$\frac{{{V_1} - {V_2}}}{{{R_2} - {R_1}}}\,(x - {R_1})$

B

$\frac{{{V_1}{R_1}({R_2} - x) + {V_2}{R_2}(x - {R_1})}}{{({R_2} - {R_1})\,x}}$

C

$\frac{{{V_1} + {V_2}}}{{{R_2} +{R_1}}}\,(x +{R_1})$

D

$\frac{{({V_1} + {V_2})}}{{({R_1} + {R_2})}}x$

Solution

${V_1} = \frac{{{Q_1}}}{{{R_1}}} + \frac{{{Q_2}}}{{{R_2}}}$…….. $(i)$

${V_2} = \frac{{{Q_2}}}{{{R_2}}} + \frac{{{Q_1}}}{{{R_2}}}$…….. $(ii)$

$V = \frac{{{Q_1}}}{x} + \frac{{{Q_2}}}{{{R_2}}} = \frac{{{Q_1}}}{x} + {V_1} – \frac{{{Q_1}}}{{{R_1}}}$ $ = {Q_1}\left( {\frac{1}{x} – \frac{1}{{{R_1}}}} \right) + {V_1} = \frac{{{Q_1}({R_1} – x)}}{{x{R_1}}} + {V_1}$ ……..$(iii)$

${V_1} – {V_2} = \frac{{{Q_1}}}{{{R_1}}} – \frac{{{Q_2}}}{{{R_2}}}$

$==>$ $({V_1} – {V_2}){R_1}{R_2} = {R_2}{Q_1} – {R_1}{Q_1}$

$==>$ ${Q_1} = \frac{{({V_1} – {V_2}){R_1}{R_2}}}{{{R_2} – {R_1}}}$ $V = \frac{{({R_1} – x)\,({V_1} – {V_2}){R_1}{R_2}}}{{x{R_1}({R_2} – {R_1})}} + {V_1}$

$==>$ $V = \frac{{{V_1}{R_1}({R_2} – x) + {V_2}{R_2}(x – {R_1})}}{{x({R_2} – {R_1})}}$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.