Charging Charge

Electricity and Magnetism Level 2

A particle of mass 9 × 1 0 − 31 k g 9\times 10^{-31} kg carrying negative charge of 1.6 × 1 0 − 19 C 1.6\times 10^{-19} C is projected horizontally between two infinite horizontal parallel plate with velocity of 1 0 6 m s − 1 10^6 ms^{-1} . The distance between the plates is d = 3 × 1 0 − 3 m d = 3\times 10^{-3} m and the particle enters at a distance of 1 0 − 3 m 10^{-3}m below the upper plate. The top and bottom plates are connected to positive and negative terminal of 30 V 30V battery respectively. Find the horizontal component v x v_x and vertical component v y v_y of velocity after the particle strikes the upper plate.

Enter your answer as 1 0 − 6 ( v x + v y ) 10^{-6} ( v_x + v_y)


The answer is 2.885.

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1 solution

We know that, F = m a = q E F = ma = \dfrac qE

a = q m E = q m V d a = \dfrac qm E = \dfrac qm \dfrac Vd

As E = − d V d r = V d {\color{#20A900}{E = -\dfrac {dV}{dr} = \dfrac Vd}}

Horizontal component remains same even after striking v x = 1 0 6 m s − 1 v_x = 10^6 ms^{-1}

For Vertical case, v y 2 = 2 a y 0 v_y^2 = 2ay_0

⇒ v y = 2 a y 0 = 2 q V y 0 m d = 2 × 1.6 × 1 0 − 19 × 30 × 1 0 − 3 9 × 1 0 − 31 × 3 × 1 0 − 3 = 1.885 × 1 0 6 m s − 1 \Rightarrow v_y = \sqrt {2ay_0} = \sqrt{\dfrac {2qVy_0}{md}}= \sqrt{\dfrac {2\times 1.6\times 10^{-19} \times 30\times 10^{-3}}{9\times 10^{-31} \times 3\times 10^{-3}}} = 1.885\times 10^6 ms^{-1}

Therefore, 1 0 − 6 ( v x + v y ) = 1.885 + 1 = 2.885 10^{-6}( v_x + v_y) = 1.885 + 1 = \boxed{2.885}

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