Question

5. A proton and an electron are in a uniform electric field with magnitude 45 V/C. Which particle has the larger magnitude acceleration, and what is the magnitude of the acceleration? (e = 1.6×10−19 C, mp = 1.67×10−27 kg, me = 9.11×10−31 kg) A. the electron, 7.9×1012 m/s2 B. the proton, 4.3×1013 m/s2 C. the proton, 4.3×109 m/s2 D. the electron, 7.9×108 m/s2 E. the electron, 7.9×1010 m/s2

6. A +2q point charge is located at the origin (x = 0), and a –3q charge is at x = +d. What is the magnitude of the electric field at x = +2d ? Assume 1 4π0 = k. A. 7kq 2d2 B. 5kq 2d2 C. 4kq d2 D. 3kq 2d2 E. 5kq d2

Answer 1

5.

Using Force balance on electron:

Fe = Fnet

q*E = m*a1

a1 = q*E/m

q = charge on electron = 1.6*10^-19 C

E = 45 V/C

m = 9.1*10^-31 kg

So,

a1 = (1.6*10^-19*45)/(9.1*10^-31) = 7.91*10^12 m/sec^2

Using Force balance on proton:

Fe = Fnet

q*E = m*a2

a2 = q*E/m

q = charge on proton = 1.6*10^-19 C

E = 45 V/C

m = mass of proton = 1.67*10^-27 kg

So,

a2 = (1.6*10^-19*45)/(1.67*10^-27) = 4.31*10^9 m/sec^2

So Electron has more acceleration than proton, electron's acceleration is 7.91*10^12 m/sec^2

Correct option is A.

6.

Electric field is given by:

E = k*Q/R^2

where direction of electric field is away from positive charge and towards the negative charge.

Net electric field at x = +2d will be:

Enet = E1 - E2

Since due to +2q charge, field will be in +ve x-axis and due to -3q charge field will be in -ve x-axis, So

Enet = k*q1/r1^2 - k*q2/r2^2

q1 = +2q & q2 = -3q

r1 = 2d & r2 = d

So

Enet = k*2q/(2d)^2 - k*3q/d^2

Enet = k*q/(2*d^2) - 3*k*q/d^2

Enet = -5*k*q/(2*d^2)

Magnitude of electric field will be:

|Enet| = 5*k*q/(2d^2)

Correct option is B.

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