An electron (q = -e = -1.602 Times 10^-19 C) is accelerated through a potential

Question

An electron (q = -e = -1.602 Times 10^-19 C) is accelerated through a potential difference of 2.00 Times 10^5 V over a distance of 20 cm. If the electric field is uniform, what is its magnitude? How much work is done on the electron? What is the force on the electron? If the mass of the electron is 9.1 Times 10^-31 kg, what is the speed of the electron? How does this compare to the speed of light (c = 3 Times 10^8 m/s)? An electric dipole of strength p = 2 Times 10^-6 C.m has its positive charge at point (0, 1 m) and its negative charge at (0, -1 m). What is the electric potential at point (2 m. 3 m)? At (-1 m, -2m)? How much work is required to move a charge of 3 mu C from the first point to the second point? A charge of 100 nC is located at the origin of a coordinate system. A second charge of 200 nC is slowly brought from a large distance to within 10 cm of the first charge. How much work is required? What is the electrical potential energy of the second charge? What was its potential energy before being moved? The electric field is given by the equation E = 2Ax i + A j, where A = 1 Times 10^4 V/m^2. How much work does the electrical field do when a charge of 2 mu C is taken from point (0, 0) along the x-axis to (1 m, 0) and parallel to the y-axis to (1m, 2 m)? How much electrical work is done when the charge is moved along a straight line from (0, 0) lo (1 m, 2 m)? What is the equation for the electric potential of this field assuming origin has a potential of zero? What is the electric potential at point (-1 m, 2 m)? Three point charges, which initially are infinitely far apart, are placed at the comers of an equilateral triangle of side a. Two of the point charges are identical and have charge q. What must be the value of the this charge if the net work in placing them at the comers of the triangle is zero? Two stationary point charges 10 nC and -20 nC are located 10 cm apart. An electron is released between them at a distance of 3 cm from the negative charge. What is the speed of the election when it point 3 cm from the positive charge?

Explanation / Answer

Here,

Potential due to a point charge is given as

V = k * q/r

r is the distance of point from charge

Now , let the magnitude of charge on dipole is q

as p = q * d

2 *10^-6 = q * 2

q = 1 *10^-6 C

a) at point (2, 3)

for the potential

V1 = 9*10^9 * 1 *10^-6 * (1/sqrt((2 -1)^2 + 3^2) - 1/sqrt( (1 + 2)^2 + 3^2)

V1 = 724.8 V

the potential at point (2 , 3) is 724.8 V

b)

at point (-1 , -2)

for the potential

V2 = 9*10^9 * 1 *10^-6 * (1/sqrt((-1 -1)^2 + 2^2) - 1/sqrt( (1 -1)^2 + 2^2)

v2 = -1318 V

the potential at this point is -1318 V

c)

work required = potential difference * charge

work required = (-1318 - 724.8) * 3 * 10^-6

work required = -6.128 *10^-3 J

the work required to move the charge is -6.128 *10^-3 J

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