Question

The magnitude of the electric field between two parallel charged plates is 200. An electron moves to the negative plate 5. 0 cm away. Two horizontal parallel plates have equally spaced vectors going from the top plate to the bottom plate labeled E and the plates are spaced a distance d apart. There is a negatively charged sphere is near the positively charged plate and a second is directly below and near the negatively charged plate. The two charged particles are connected by a dashed line vector point from the top to the bottom one. Find the electric potential difference and the work. Recall the charge of an electron is 1. 602 × 10–19 C. ΔV = V Round work to one decimal. W = × 10–18 J.

Answer 1

The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

What is an electric field?

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The given data in the problem is given by;

E is the electric field = (200 N/C)

d is the distance = 5.0 cm.=0.05 m

Q is the charge of electrons= 1.602 x 10^-19 C

The formula for electric potential  is given by;

$\rm V=Ed$

$\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V= 10 \frac{Nm}{C} = 10 \frac{J}{C} = 10 V.$

The work is defined as the product of the potential difference and charge of an electron.

$\rm W= 10 \times 1.602 x 10^{-19} \\\\\ \rm W= 1.6 x 10^{-18 }J$

Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

To learn more about the electric field refer to the link;

brainly.com/question/15071884