Question

A charged particle of mass 4 gm and having a charge
+1 pc is fired from a point with speed 10 m/s where
electric potential is 4 x 10 V in the direction where
electric potential is decreasing, find the potential ata
point (in x10 volts) where its speed becomes 10 m/s
Express your answer in multiple of 10. 4​

Answer 1

Answer:

sorry I don't know the answer

sorry mate

Answer 2

"It seems this is what you are looking for"

A charged particle of mass 4gm and having a charge +1μC is fired from a point with speed 10ms⁻¹ ,where electric potential is 4×10⁵V in the direction, where electric potential is decreasing, find the potential at a point (in volts) where its speed becomes 10√2ms⁻¹ .Express your answer in multiple of 10⁵:

Answer:

The potential of charged particle will be 2×10⁵V  where its speed becomes 10√2ms⁻¹.

Step-by-step explanation:

Initial electric potential $V_{i}=4*10^{5}V$

mass of particle $m=4g=4*10^{-3}Kg$

$q=1\mu C=10^{-6}C$

By applying the work -energy theorem:

Work done by electrostatic forces = change in kinetic energy

$q(v_{i}-v_{f})=\frac{1}{2}mv^{2}_{f} -\frac{1}{2}mv^{2}_{i}$

$10^{-6}(v_{i}-v_{f})=\frac{1}{2}*4*10^{-3}[(10\sqrt{2})^{2} -(10)^{2}]$

$10^{-6}(v_{i}-v_{f})=2*10^{-3}(200)$

$(v_{i}-v_{f})=2*10^{5}$

$(4*10^{5}-v_{f})=2*10^{5}\\v_{f}=4*10^{5}-2*10^{5}\\v_{f}=2*10^{5}V$

Therefore the required electric potential would be 2×10⁵V.