A proton released from rest at a distance of 5 cm from a stationary 25 mu C char

Question

A proton released from rest at a distance of 5 cm from a stationary 25 mu C charge. What is the speed of the proton when it reached a point which is 10 cm from this charge? [Mass of a proton is 1.67 times 10^27 kg, charge e = 1.6 times 10^29 C) What is the strength of the electric field between two big parallel conducting plates separated by 2 mm if the potential difference between these plates is 3000 V? Two capacitors C_1 = 15mu F and C_2 = 30 mu F are connected in series with a 40 V battery. Find the equivalent capacitance. Find the charges Q_1 and Q_3 on these capacitors and the voltages V_1 and V_2 across each capacitor.

Explanation / Answer

1) let

Q = 25 micro C

at r1 = 5 cm = 0.05 m,

V1 = k*Q/r1

= 9*10^9*25*10^-6/0.05

= 4.5*10^6 V

at r2 = 10 cm = 0.1 m

V2 = k*Q/r2

= 9*10^9*25*10^-6/0.1

= 2.25*10^6 V

Workdone on proton = gain in kinetic energy

q*(V1 - V2) = (1/2)*m*v^2

==> v = sqrt(2*q*(V1 - V2)/m)

= sqrt(2*1.6*10^-19*(4.5 - 2.25)*10^6/(1.67*10^-27))

= 2.08*10^7 m/s

2) E = delta_V/d

= 3000/(2*10^-3)

= 1.5*10^6 N/c

3)

a) Ceq = C1*C2/(C1+C2)

= 15*30/(15 + 30)

= 10 micro F

b)
Q1 = Q2 = Ceq*V

= 10*40

= 400 micro C or 4*10^-4 C

V1 = Q1/C1 = 400/15 = 26.7 V

V2 = Q2/C2 = 400/30 = 13.3 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.