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(B) Since the spheres are joined by a metal wire,their potentials will be the same.Let the radius of the smaller sphere be $r_1$ and the radius of the

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quarter of that near the smaller sphere

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(B) Since the spheres are joined by a metal wire,their potentials will be the same.Let the radius of the smaller sphere be $r_1$ and the radius of the larger sphere be $r_2 = 4r_1$.Since the potentials are equal,$V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2}$.Therefore,$\frac{q_2}{q_1} = \frac{r_2}{r_1} = 4$.The electric field near the surface of the spheres is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}$.Thus,$\frac{E_2}{E_1} = \frac{q_2}{q_1} \cdot \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{r_2}{r_1}\right) \cdot \left(\frac{r_1}{r_2}\right)^2 = \frac{r_1}{r_2}$.Substituting $r_2 = 4r_1$,we get $\frac{E_2}{E_1} = \frac{r_1}{4r_1} = \frac{1}{4}$.So,the electric field near the larger sphere is a quarter of that near the smaller sphere.

Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere,the electric field near the larger sphere is

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