Question

16 micro columb and 4 micro charges are separated by 24 cm where should be the third charge 5 nanocoulomb it will be placed so that it will be in equilibrium

Answer 1

Answer:

Given:

16 and 4 micro C charge are separated by 24 cm.

To find:

Position of 5 nano C so as to remain in equilibrium.

Concept:

For the charge to remain in equilibrium , it has to be placed in between the 16 and 4 micro C charges. And also it has to be placed a little nearer to the 4 micro C charge.

Calculation:

Let the equilibrium point be x cm from 16 micro C. Hence the other distance will be (24 - x)

$F1 = F2$

$\implies \: \frac{1}{4\pi \epsilon} \frac{(16 \times {10}^{ - 6}) \times (5 \times {10}^{ - 9} )}{ {x}^{2} } = \frac{1}{4\pi \epsilon} \frac{(4 \times {10}^{ - 6}) \times (5 \times {10}^{ - 9} )}{ {(24 - x)}^{2} }$

Cutting all the necessary terms :

$\implies \: \dfrac{16 \times 5}{ {x}^{2} } = \dfrac{4 \times 5}{ {(24 - x)}^{2} }$

$\implies \: \dfrac{16}{ {x}^{2} } = \dfrac{4}{ {(24 - x)}^{2} }$

$\implies \: \dfrac{4}{x} = \dfrac{2}{(24 - x)}$

$\implies \: 96 - 4x = 2x$

$\implies \: 6x = 96$

$\implies \: x = 16 \: cm$

So the charge is located 16 cm from 16 micro C charge and 8 cm from 4 micro C.

Answer 2

$\underline{ \boxed{ \bold{ \purple{ \huge{Answer\::-}}}}} \\ \\ \star \rm \: \red{Given} \\ \\ \mapsto \rm \: 16 \mu{C} \: and \: 4 \mu{C} \: charges \: are \: separated \\ \rm \: by \: 24cm...another \: charge \: of \: 5nC \: is \: placed \\ \rm \: between \: two \: static \: charges... \\ \\ \star \rm \: \red{To \: Find} \\ \\ \mapsto \rm \: where \: should \: be \: third \: charge \: placed \\ \rm \: so \: that \: whole \: system \: remains \: into \: eq... \\ \\ \star \rm \: \red{Concept} \\ \\ \mapsto \rm \: here \: all \: charges \: are \: positive \: in \: nature \\ \rm \: so \: we \: should \: place \: third \: charge \: at \: between \\ \rm \: two \: charges \: for \: condition \: of \: eq... \\ \\ \star \rm \: \red{Assumption} \\ \\ \mapsto \rm \: let \: distance \: between \: 16 \mu{C} \: and \: 5nC\: charge \\ \rm \: is \: x \: cm \: and \: value \: of \: pushing \: force \: is \: F_1... \\ \\ \mapsto \rm \: so\: distance \: between \: 4 \mu{C} \: and \: 5nC \: charge \\ \rm \: will \: be \: (24 - x) \: cm \: and \: value \: of \: pushing \\ \rm \: type \: force \: will \: be \: F_2 \\ \\ \star \rm \: \red{ Formula} \\ \\ \mapsto \rm \: for \: condition \: of \: equilibrium... \\ \\ \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \purple{ \rm{ \huge{\bold{F_1 = F_2}}}}}}\\ \\ \star \rm \: \red{Calculation} \\ \\ \leadsto \rm \: F_1 = F_2 \\ \\ \leadsto \rm \: \frac{k(16 \mu{C})(5nC)}{ {x}^{2} } = \frac{k(5nC)(4 \mu{C})}{ {(24 - x)}^{2} } \\ \\ \dagger \rm \: cutting \: down \: same \: values... \: \dagger \\ \\ \leadsto \rm \: \frac{80}{ {x}^{2} } = \frac{20}{ {(24 - x)}^{2} } \\ \\ \leadsto \rm \: 4 = \frac{ {x}^{2} }{ {(24 - x)}^{2} } \\ \\ \leadsto \rm \: x = 2(24 - x) \\ \\ \leadsto \rm \: 3x = 48 \\ \\ \leadsto \rm \: \blue{x = 16 \: cm} \\ \\ \dagger \rm \: \: for \: eq. \: position \: third \: charge \: should \: be \: \orange{16cm} \\ \rm \: apart \: from \: \orange{16 \mu{C}} \: charge \: and \: \pink{8 \: cm} \: apart \: from \\ \rm \: \pink{4 \mu{C} }\: charge... \: \: \dagger \\ \\ \star \rm \: \red{Diagram} \\ \\ \mapsto \rm \: see \: the \: attached \: image...$