Question

1.) A particle is uncharged and is thrown vertically upward from ground level with a speed of 20.1 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 27.3 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge −q. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

2.) A moving particle encounters an external electric field that decreases its kinetic energy from 9190 eV to 6350 eV as the particle moves from position A to position B. The electric potential at A is -51.0 V, and the electric potential at B is +26.7 V. Determine the charge of the particle. Include the algebraic sign (+ or −) with your answer.

3.) Identical +2.22 µC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is 0 V? .

4) A charge of -2.75 µC is fixed in place. From a horizontal distance of 0.0465 m, a particle of mass 6.85 10-3 kg and charge -7.40 µC is fired with an initial speed of 59.5 m/s directly toward the fixed charge. How far does the particle travel before its speed is zero?

5.) The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.062 V exists across the membrane. The thickness of the cell membrane is 8.80 10-9 m. What is the magnitude of the electric field in the membrane?

6.) A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.65 mm. When an electric spark jumps between them, the magnitude of the electric field is 5.30 107 V/m. What is the magnitude of the potential difference ΔV between the conductors?

Answer 1

1. given
intiial speed, u = 20.1 m/s
maximum height h
charge given afterwards, q
speed, v = 27.3 m/s
maximum height = h
charge, -q
speed, v' = ?
for same height = h

let potneital difference between the same height difference be V
then
from energy balance
for mass of particel m
0.5mu^2 = mgh
0.5mv^2 = mgh + qV
0.5mv'^2 = mgh - qV

0.5m(v'^2 + v^2) = 2mgh = mu^2
v'^2 + v^2 = 2u^2
v' = sqroot(2u^2 - v^2) = 7.92022 m/s

2. given
Va = -51 V
Vb = 26.7 V
dE = 9190 - 6350 = 2840 eV
dE = 2840 eV = 2840*1.6*10^-19 = q(Vb - Va)
q = 5.84813*10^-18 C

3. given charge q = 2.22 uC are on two adjacent corners of a square of side a
charge on third corner = Q
electric potential on the 4th corner = 0 V
hence
kq/sqrt(2)*a + kq/a + kQ/a = 0
q/sqrt(2) + q = -Q
Q = -q(1 + 1/sqrt(2)) = -3.789777 uC

4. q = -2.75 uC
x = 0.0465 m
m = 6.85*10^-3 kg
q' = -7.4 u C
u = 59.5 m/s
distance travelled = d
hence
from conservation of energy
0.5mu^2 = kqq'/(x - d) - kqq'/(x)
66.35196 = (1/0.0465 - d) - 1/0.0465
d = 0.03511791
m