Physics, asked by sandhyaprakash079, 10 months ago

+3q and -4q charges are placed along x axis 1m apart. Where must be +q charge placed so that +q charge will be in equilibrium.​

Answers

Answered by Anonymous
33

Solution

  • The charge should be placed in proximity to +3Q

Given

Two Charges of magnitude + 3Q and -4Q are placed one meters apart along the positive x - axis

To finD

A charge Q can be placed so that the former would be in equilibrium

For equilibrium,

The net charge of the system should be zero

By symmetry,

The +Q charge should be placed near +3Q so that this would balance out the -4Q charge and the net charge remains the same

Answered by HappiestWriter012
5

The charge +q shouldn't be placed between the charges +3q & -4q as the direction of Force of repulsion by +3q & - 4q will be along the same direction, and there will be a net force.

So we should place the charge +q at some distance from +3q measured towards - X axis, but not between the charges.

Given Distance between the charges is 1m.

( Measured towards -ve X axis) Let the distance from +q to +3q be x

Then, (Measured towards - ve X axis) The distance between +q & - 4q = 1+x

By Coulomb's law,

Force on +q due to +3q is given by,

F =  \frac{k(q)(3q)}{ {x}^{2} }  =  \frac{3k {q}^{2} }{  {x}^{2}  }

Force on +q due to - 4q is given by,

F =  \frac{k(q)( - 4q)}{(1 + x) ^{2} }  =  \frac{ - 4k {q}^{2} }{(1 + x) ^{2} }

For the charge to be in equilibrium

Total force on the charge = 0

 \implies \dfrac{3k {q}^{2} }{ {x}^{2} }  +  \dfrac{ - 4k {q}^{2} }{(1 + x) ^{2} }  = 0 \\  \\ \implies \dfrac{3k {q}^{2} }{ {x}^{2} }   =   \dfrac{  4k {q}^{2} }{(1 + x) ^{2} }   \\  \\  \implies \:  \frac{3}{ {x}^{2} }  =  \frac{4}{ {(1 + x)}^{2} }  \\  \\  \implies \:  \pm \:  \frac{ \sqrt{3} }{2}  =  \frac{x}{1 + x}  \\  \\

Considering positive value

⇒√3 ( 1 + x) = 2x

⇒ √3 = (2 - √3) x

⇒ √3 ( 2 + √3) = x

⇒3 + 2√3 = x ___ (1)

Considering negative value

-√3 ( 1 + x) = 2x

⇒- √3 = (2 + √3) x

⇒ - √3 ( 2 - √3) = x

⇒ 3 - 2√3 = x _____(2)

Since (2) is a negative value, It is ignored.

Therefore, The Charge +q should be placed at a distance 3 + 2√3 from the charge +3q measured in the direction of -ve X axis.

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