As shown in figure, a dust particle with mass m=5.0 × 10-9 kg and charge q0=2.0 nC starts from rest at point a and moves in a straight line to point b. What is its speed v at point b ?

Question

As shown in figure, a dust particle with mass m=5.0 × 10-9 kg and charge q0=2.0 nC starts from rest at point a and moves in a straight line to point b. What is its speed v at point b ? (A) 26 ms -1 (B) 34 ms -1 (C) 46 ms -1 (D) 14 ms -1

Tardigrade Admin · Accepted Answer

Apply conservation of mechanical energy between points

a

and

b

(K E + PE)_{a} = (K E + PE)_{b}

0 + \frac{k ( 3 \times 1 0 ^{- 9} ) q _{0}}{0.01} - \frac{k ( 3 \times 1 0 ^{- 9} ) q _{0}}{0.02}

= \frac{1}{2} m v^{2} + \frac{k ( 3 \times 1 0 ^{- 9} ) q _{0}}{0.02} - \frac{k ( 3 \times 1 0 ^{- 9} ) q _{0}}{0.01}

Put the values and get

v = 1215 = 46 m s^{- 1}

Q. As shown in figure, a dust particle with mass $m = 5.0$ $\times 1 0^{- 9} k g$ and charge $q_{0} = 2.0 n C$ starts from rest at point $a$ and moves in a straight line to point $b$ . What is its speed $v$ at point $b$ ?

$26 m s^{- 1}$

$34 m s^{- 1}$

$46 m s^{- 1}$

$14 m s^{- 1}$

Q. As shown in figure, a dust particle with mass m=5.0 ×10−9kg and charge q0​=2.0nC starts from rest at point a and moves in a straight line to point b. What is its speed v at point b ?

26ms−1

34ms−1

46ms−1

14ms−1

Apply conservation of mechanical energy between points a and b (KE+PE)a​=(KE+PE)b​ 0+0.01k(3×10−9)q0​​−0.02k(3×10−9)q0​​ =21​mv2+0.02k(3×10−9)q0​​−0.01k(3×10−9)q0​​ Put the values and get v=1215​=46ms−1

Q. As shown in figure, a dust particle with mass $m = 5.0$ $\times 1 0^{- 9} k g$ and charge $q_{0} = 2.0 n C$ starts from rest at point $a$ and moves in a straight line to point $b$ . What is its speed $v$ at point $b$ ?

$26 m s^{- 1}$

$34 m s^{- 1}$

$46 m s^{- 1}$

$14 m s^{- 1}$