Question

The potential in the region of space near the point P (-2,4,6) is V=80x²+60y²V(a) Find out the electric Field vector in the region(b) Find out the Electric field vector at point P(c) What is the value of potential at point P​

Answer 1

Answer:

(b) (320 i - 480 j) ; (c) 1280 volt

Explanation:

(b) E = –[(dv/dx) i + (dv/dy) j + (dv/dz) k] = –[160x i + 120y j]

at point P(-2,4,6) E = (320 i – 480j)

(c) at P value of potential, V = 80×(-2)² + 60×4² = 1280 V =1.28 kV

Answer 2

Given:

• Potential function is V = 80x² + 60y².

To find:

• Field vector ?
• Field intensity value at point P ?
• Potential at point P ?

Calculation:

Field intensity vector is :

$\vec{E} = - \bigg \{ \dfrac{ \partial V}{ \partial x} \hat{i} + \dfrac{ \partial V}{ \partial y} \hat{j} \bigg \}$

$\implies \vec{E} = - \bigg \{ \dfrac{ \partial (80 {x}^{2} + 60 {y}^{2} ) }{ \partial x} \hat{i} + \dfrac{ \partial (80 {x}^{2} + 60 {y}^{2} )}{ \partial y} \hat{j} \bigg \}$

$\boxed{ \implies \vec{E} = - \bigg \{ 160x \: \hat{i} + 120y \hat{j} \bigg \}}$

At point P , the value will be :

$\implies \vec{E} \bigg| _{P} = - \bigg \{ 160( - 2) \: \hat{i} + 120(4) \hat{j} \bigg \}$

$\boxed{\implies \vec{E} \bigg| _{P} = 320 \: \hat{i} - 480 \hat{j}}$

Value of potential at point P :

$V = 80 {x}^{2} + 60 {y}^{2}$

$\implies V = 80(4) + 60(16)$

$\implies V = 320 + 960$

$\implies V =1280 \: volts$

Hope It Helps