a) At r=45 cm from the center (r>R):
"V=\\frac{kq}{r}=\\frac{9\u00b710^9\u00b74\u00b710^{-6}}{0.45}=79.9\u00b710^3\\text{ V}."b) and c) At the surface and inside for "r\\leq R" the sphere the field is zero and the potential is
"V=\\frac{kq}{R}=\\frac{9\u00b710^9\u00b74\u00b710^{-6}}{0.3}=120\u00b710^3\\text{ V}."
Comments
Leave a comment