Question

the electric potential in a region is given by v equal to 3 X square + 4 x an electron is placed at the origin will experience
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Answer 1

Answer:

Given:

Electric Potential in a region is given as :

$V = 3 {x}^{2} + 4x$

To find:

Force experienced by an electron placed at origin.

Concept:

Electric Potential is an Scalar Index representing the field due to a Source.

Mathematically , we can perform partial derivative to get the field intensity experienced.

Let Potential be V , and Field Intensity be E.

$\sf{ \huge{ E = - (\dfrac{ \delta V}{ \delta x})}}$

Calculation:

$\sf{ E = - \bigg\{\dfrac{ \delta (3 {x}^{2} + 4x) }{ \delta x} \bigg \}}$

$\sf{ E = - (6x + 4)}$

At origin ( 0,0) , the Field Intensity will be :

$\sf{ E = - \{(6 \times 0) + 4 \}}$

$\sf{ E = - 4 \: V {m}^{ - 1} }$

We know that force is given as the product of Electric Field Intensity and Charge :

$\sf{Force = - 4 \times ( - 1.6 \times {10}^{ - 19} )}$

$\sf{Force = 6.4 \times {10}^{ - 19} N }$

Final answer :

$\boxed{ \huge{ \green{\sf{Force = 6.4 \times {10}^{ - 19} N }}}}$

Answer 2

$\underline{ \boxed{ \bold{ \mathfrak{ \purple{ \huge{Answer}}}}}}$

Given :

$\rm \: electric \: potential \: in \: a \: region \: is \: \\ \rm \: given \: by \: \red{3 {x}^{2} + 4x}$

To Find :

$\rm \: force \: experienced \: by \: electron \: at \: origin$

Formula :

$\rm \: Electric \: potential \: represent \: work \: done \\ \rm \: to \: bring \: a \: charge \: from \: infinite \: to \: a \\ \rm \: point \: which \: is \: situated \: in \: electric \: field \\ \rm \: of \: another \: charged \: particle...We \: can \: get \\ \rm \: electric \: field \: by \: partial \: differenciation \: of \: \\ \rm \: potential \: w.r.t. \: distance... \\ \\ \rm \: Mathematically \implies \: \underline{ \boxed{ \bold{ \rm{ \pink{E = - \frac{ \delta{V}}{ \delta{x}} }}}}} \\ \\ \rm \: Electric \: field \: is \: defiend \: as \: force \: experienced\: by \\ \rm \: unit \: charge \: particle. \\ \\ \rm \: mathematically \implies \underline{ \boxed{ \bold{ \rm{ \blue{F = qE}}}}}$

Calculation :

$\mapsto \rm \: E = - \frac{ \delta({3 {x}^{2} + 4x })}{ \delta{x}} = - (6x + 4) \\ \\ \mapsto \rm \: Field \: intensity \: at \: origin = \green{- 4 \: \frac{N}{C}} \: ( \because{x = 0}) \\ \\ \mapsto \rm \: Force \: experienced \: by \: electron = q \times e \\ \\ \mapsto \rm \: F =( - 1.6 \times {10}^{ - 19} ) \times ( - 4) \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{F = 6.4 \times {10}^{ - 19} \: N}}}}} \: \purple{ \star}$