Question

a positively charged sphere with a charge of 2.6 x 10^-12 c and a mass of 3.0 x 10^-15 kg is placed near the positive plate of a set of parallel plates. the electric potential between the plates is 2.0 x 10^3 v. when released, the positively charged particle accelerates from rest towards the negative plate. calculate the spheres speed the instant before it hits the negative plate. a. 1.9 x 10^3 m/s b. 5.2 x 10^4 m/s c 4.3 x 10^5 m/s d. 3.5 x 10^6 m/s

Explanation:

Step1: Identify the energy transformation

The electric potential energy of the charged sphere is converted into kinetic energy as it moves from the positive plate to the negative plate. The formula for electric potential energy is \( U = qV \), and the formula for kinetic energy is \( K = \frac{1}{2}mv^2 \). By conservation of energy, \( qV=\frac{1}{2}mv^2 \).

Step2: Rearrange the formula to solve for velocity \( v \)

From \( qV=\frac{1}{2}mv^2 \), we can solve for \( v \):

$$ v=\sqrt{\frac{2qV}{m}} $$

Step3: Substitute the given values

Given \( q = 2.6\times 10^{- 12}\space C \), \( V=2.0\times 10^{3}\space V \), and \( m = 3.0\times 10^{-15}\space kg \).
Substitute these values into the formula:

$$ v=\sqrt{\frac{2\times(2.6\times 10^{-12}\space C)\times(2.0\times 10^{3}\space V)}{3.0\times 10^{-15}\space kg}} $$

First, calculate the numerator:

$$ 2\times(2.6\times 10^{-12})\times(2.0\times 10^{3})=2\times2.6\times2.0\times 10^{-12 + 3}=10.4\times 10^{-9}=1.04\times 10^{-8} $$

Then divide by the mass:

$$ \frac{1.04\times 10^{-8}}{3.0\times 10^{-15}}=\frac{1.04}{3.0}\times 10^{-8+15}\approx0.3467\times 10^{7}=3.467\times 10^{6} $$

Now take the square root:

$$ v=\sqrt{3.467\times 10^{6}}\approx1.86\times 10^{3}\space m/s\approx1.9\times 10^{3}\space m/s $$

Answer:

a. \( 1.9\times 10^{3}\space m/s \)