# Near the surface of the earth, there is an electric field of about 150 V/m which points downward....

## Question:

Near the surface of the earth, there is an electric field of about 150 V/m which points downward. Two identical balls with mass m = 0.520 kg are dropped from a height of 2.00 m, but one of the balls is positively charged with q{eq}_1 {/eq} = 600 {eq}\mu {/eq}C, and the second is negatively charged with q{eq}_2 {/eq} = -600 {eq}\mu {/eq}C. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

## Conservation of Energy:

Conservation of energy describes a system with constant total energy. The total energy before an interaction will be the same after an interaction. Usually kinetic energy will be converted entirely to potential energy, making the energy of the system constant.

Conservation of energy is given as:

{eq}\frac{1}{2}mv^2 = mgh + qEh {/eq}

where:

• m is the mass
• q is the charge
• E is the electric field
• g is the acceleration due to gravity

For the first ball, the speed is given as:

{eq}\frac{1}{2}(0.520)v^2 = (0.520)(9.8)(2.0) + (600\times10^{-6})(150)(2.00) \\ v = 6.31 \ m/s {/eq}

The speed of the second ball is:

{eq}\frac{1}{2}(0.520)v^2 = (0.520)(9.8)(2.0) - (600\times10^{-6})(150)(2.00) \\ v = 6.20 \ m/s {/eq}

The difference in speed is given as:

{eq}\Delta v = 6.31-3.20 = 0.11 \ m/s {/eq}