Question

a charged particle of 2mc is released from rest in electric field e=2x10^3n/c .find the magnitude and direction of force.

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Answer 1

Answer:

$\\\tt 4 \times {10}^{ - 3} N$

Explanation:

Force on a charge in an electric feild is given by qE

where q = charge

E = electric feild

Using given data F =

$\\\tt(2 \times {10}^{ - 6} )(2 \times {10}^{3} ) N \\ \\\tt = 4 \times {10}^{ - 3} N$

Since electric feild and charge both are +ve it's direction will be away from the feild.

Answer 2

Given : A charged particle of 2mc ( q ) in Electrical feild which is E = 2 × 10³ N / C ( E ) .

Exigency To Find : Magnitude and direction of force.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

• Electric feild which is E = 2 × 10³ N / C ( E ) .
• The Charge is 2 mc ( or micro columb )

⠀⠀⠀⠀⠀Converting the charge from mc ( or micro coulomb ) to c ( or Coulomb ) :

$:\implies \sf Charge \:(\:q\:)\: = \: 2 \: mC \: \\\\ :\implies \sf Charge \:(\:q\:)\: = \: 2 \times 10^{-6} \: C \:\qquad \:\bigg\lgroup \sf{ 1 \: mC \:(or \:micro\: coulomb)\:=\:1 \times 10^{-6}\: C\: ( or \:coulomb) }\bigg\rgroup\\\\ :\implies \sf Charge \:(\:q\:)\: = \: 2 \times 10^{-6} \: C \: \\\\ \qquad \therefore \pmb{\underline{\purple{\:Charge\:(\: q\:)\:=\:\: 2 \:\times 10^{-6} \: C }} }\:\:\bigstar$

⠀⠀⠀⠀⠀Finding Magnitude of Force :

$\dag\:\:\frak{ As,\:We\:know\:that\::}\\\\\qquad\bigstar\:\bf Formula\:for \:Force\: \::$

$\qquad \dag\:\:\bigg\lgroup \sf{ \qquad Force \:(\: F\:)\:=\:\:Eq \qquad}\bigg\rgroup$

⠀⠀⠀⠀⠀Here , f is the magnitude of Force , q is the charge & E is the Electric feild .

$\dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\:Eq \:\:\\\\ \qquad \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\ \dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\:Eq \:\:\\\\ \dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\bigg(\: 2 \:\times 10^{-6} \: \:\bigg)\:C \:\times \bigg( \:2 \times 10^3 \:\bigg) N / C \:\:\\\\ \dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\bigg(\: 2 \:\times 10^{-6} \: \:\bigg)\:\cancel {C }\:\times \bigg( \:2 \times 10^3 \:\bigg) N/\cancel {C}\:\:\\\\ \dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\bigg(\: 2 \:\times 10^{-6} \: \:\bigg)\:\times \bigg( \:2 \times 10^3 \:\bigg) N \:\:\\\\ \dashrightarrow \sf\:\: Force \:(\: F\:)\:=\:\bigg(\: 4 \:\times 10^{-3} \: \:\bigg)\:N \:\:\\\\ \qquad \therefore \pmb{\underline{\purple{\:Force \:(\: F\:)\:=\:\: 4 \:\times 10^{-3} \: N }} }\:\:\bigstar$

⠀⠀⠀⠀⠀Here , F denotes to magnitude of force which is 4 × $10^{-3}$ N .

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf Hence, \:Magnitude \:of\:force \:is\:\bf 4 \times 10^{-3} \:\: N \: }}$

Since , The Charge and Electric feild both are in Positive " + ve " so it will move radially outward .