Question

the potential of an electrostatic field is given by V= 2x² . Determine the field strength at the point (2m) .​

Answer 1

Answer:

E= 8 N/C

Explanation:

Ex = - dv/dx = -4x

at x= 2 Ex = -8 N/C

Answer 2

Solution :

⏭ Given:

✏ The potential of an electric field is given by V = $\sf2x^2$

⏭ To Find:

✏ The electric field strength at the point 2m

⏭ Concept:

✏ Electric field strength can be calculate by partial differentiation of electrical potential.

⏭ Calculation:

✏ Relation between electric field strength and elctrical potential is given by

$\bigstar \: \underline{ \boxed{ \bold{ \sf{ \pink{E = - \dfrac{ \delta{V}}{ \delta{x}}}}}}} \: \bigstar$

⏭ Terms indication:

✏ E denotes electric field intensity

✏ V denotes electrical potential

⏭ Calculation:

$\dashrightarrow \sf \: E = - \dfrac{ \delta{V}}{ \delta{x}} \\ \\ \dashrightarrow \sf \: E = - \frac{ \delta(2 {x}^{2}) }{ \delta{x}} \\ \\ \dashrightarrow \sf \: E = - 4x \\ \\ \red{\dag \sf \: putting \: x = 2} \\ \\ \dashrightarrow \sf \: E = - ( 4 \times 2) \\ \\ \gray\bigstar\: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{E = - 8 \: N {C}^{ - 1}}}}}}} \: \gray{ \bigstar}$