2 oops edited Oct 25 '13 at 13:51 John Rennie 287k4646 gold badges588588 silver badges834834 bronze badges The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is: $$F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2}$$ and the field strength at the surface of the second sphere is: $$F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2}$$ Lets take the ratio $$Q_1/Q_2$$$$F_1/F_2$$ to see which is greater. The constants cancel to give us: $$\frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}}$$ and I'm going to rewrite this slightly to make it obvious how you use your equality $$Q_1/r_1 = Q_2/r_2$$: $$\frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}}$$ Because $$Q_1/r_1 = Q_2/r_2$$ we can cancel them on the top and bottom of the fraction and we're left with: $$\frac{F_1}{F_2} = \frac{r_2}{r_1}$$ and because $$r_2 < r_1$$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1. The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is: $$F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2}$$ and the field strength at the surface of the second sphere is: $$F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2}$$ Lets take the ratio $$Q_1/Q_2$$ to see which is greater. The constants cancel to give us: $$\frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}}$$ and I'm going to rewrite this slightly to make it obvious how you use your equality $$Q_1/r_1 = Q_2/r_2$$: $$\frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}}$$ Because $$Q_1/r_1 = Q_2/r_2$$ we can cancel them on the top and bottom of the fraction and we're left with: $$\frac{F_1}{F_2} = \frac{r_2}{r_1}$$ and because $$r_2 < r_1$$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1. The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is: $$F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2}$$ and the field strength at the surface of the second sphere is: $$F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2}$$ Lets take the ratio $$F_1/F_2$$ to see which is greater. The constants cancel to give us: $$\frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}}$$ and I'm going to rewrite this slightly to make it obvious how you use your equality $$Q_1/r_1 = Q_2/r_2$$: $$\frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}}$$ Because $$Q_1/r_1 = Q_2/r_2$$ we can cancel them on the top and bottom of the fraction and we're left with: $$\frac{F_1}{F_2} = \frac{r_2}{r_1}$$ and because $$r_2 < r_1$$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1. 1 answered Oct 25 '13 at 12:23 John Rennie 287k4646 gold badges588588 silver badges834834 bronze badges The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is: $$F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2}$$ and the field strength at the surface of the second sphere is: $$F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2}$$ Lets take the ratio $$Q_1/Q_2$$ to see which is greater. The constants cancel to give us: $$\frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}}$$ and I'm going to rewrite this slightly to make it obvious how you use your equality $$Q_1/r_1 = Q_2/r_2$$: $$\frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}}$$ Because $$Q_1/r_1 = Q_2/r_2$$ we can cancel them on the top and bottom of the fraction and we're left with: $$\frac{F_1}{F_2} = \frac{r_2}{r_1}$$ and because $$r_2 < r_1$$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1.