Two spheres of equal radii have charges q and 3q the ratio of their surface charge densities is

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Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is:

\(\begin{array}{l}\frac{3Q_{2}}{4\pi \varepsilon _{0}R}\end{array} \)

\(\begin{array}{l}\frac{3Q_{1}}{4\pi \varepsilon _{0}R}\end{array} \)

\(\begin{array}{l}\frac{3Q_{1}}{16\pi \varepsilon _{0}R}\end{array} \)

\(\begin{array}{l}\frac{Q_{1}}{4\pi \varepsilon _{0}R}\end{array} \)

Answer: 3

\(\begin{array}{l}\sigma =\frac{Q_{1}}{4\pi R^{2}}=\frac{Q_{2}}{4\pi 16R^{2}}\end{array} \)

16Q1=Q2

\(\begin{array}{l}V_{R}-V_{4R}=\frac{KQ_{1}}{R}+\frac{KQ_{2}}{4R}-\frac{KQ_{1}}{4R}-\frac{KQ_{2}}{4R}\end{array} \)

\(\begin{array}{l}=\frac{3KQ_{1}}{4R}=\frac{3Q_{1}}{16\pi \varepsilon _{0}R}\end{array} \)

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Text Solution

Solution : Let the charge densities of two spheres be ` sigma_1 = sigma_2 = sigma=(q)/(A)`. For given two sphere , the charge density is the same. `:. sigma ` = constant `=(q)/(A) implies q prop A to` Ratio of charges `q_1 : q_2 = 1:4` `:. (1) implies (q_1)/(q_2) = (A_1)/(A_2) = (1)/(4)` `(4 pi r_1^(2))/( 4pi r_2^(2)) = (1)/(4)` ` (r_1)/(r_2) = sqrt((1)/(4)) = (1)/(2)` `:.` Ratio of their volumes `=(v_1)/(v_2)=((4//3) pi r_1^(3))/((4//3)pi r_2^(3))` `(v_1)/(v_2) = ((r_1)/(r_2))^(3) = ((1)/(2))^(3)` `v_1 : v_2 =1 :8` `:.` Ratio of their volumes = 1:8