Two spherical conductors of radii $R_{1}$ and $R_{2}$ are separated by a distance much larger than the radius of either sphere. The spheres are connected by a conducting wire as shown in Fig. 3.128. If the charges on the spheres in equilibrium are $q_{1}$ and $q_{2}$ , respectively, what is the ratio of the field strength at the surfaces of the spheres ?

R_{2} / R_{1}

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R_{2}^{2} / R_{1}^{1}

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R_{1} / R_{2}

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R_{1}^{2} / R_{2}^{2}

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Solution

The correct option is A $R_{2} / R_{1}$
Their potential will be same, i.e, $V_{1} = V_{2}$
or $\frac{k q_{1}}{R_{1}} = \frac{k q_{2}}{R_{2}} o r \frac{q_{1}}{q_{2}} = \frac{R_{1}}{R_{2}}$ (i)
or $\frac{E_{1}}{E_{2}} = \frac{k q_{1} / R_{1}^{2}}{k q_{2} / R_{2}^{2}} = \frac{q_{1}}{q_{2}} {(\frac{R_{2}}{R_{1}})}^{2}$ (ii)
now using equation (i) and(ii)
$\frac{E_{1}}{E_{2}} = \frac{R_{1}}{R_{2}} {(\frac{R_{2}}{R_{1}})}^{2} = \frac{R_{2}}{R_{1}}$

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The correct option is A R2/R1Their potential will be same, i.e, V1=V2or kq1R1=kq2R2 or q1q2=R1R2(i)or E1E2=kq1/R21kq2/R22=q1q2(R2R1)2(ii)now using equation (i) and(ii)E1E2=R1R2(R2R1)2=R2R1