Electric field is space is given by E- -E0-a x - E0 -10 N-C- a - 1 metre- find potential of point 4-3 assuming potential at origin is 100 volt

DV = E⃗. d⃗r⃗ = {Ex î+ Ey ĵ}. {dx î+ dy ĵ} ∫ dV = Ex dx = E_0 x dx here,a=1 Δ V = ∫ dv = ∫_0^4 E_0 x dx = E_0 x^2/2 = 5x^2 = 5 (16) V 100 = 80 V =20 ...

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Byju's Answer

Standard XIII

Physics

Potential Gradient and Relation between Electric Field and Potential

Electric fiel...

Question

Electric field is space is given by $\to E = \frac{E_{0}}{a} x^j (E_{0} = 10 N / C$ , a = 1 metre), find potential of point (4,3) (assuming potential at origin is 100 volt)

10 volt

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25 volt

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20 volt

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5 volt

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Solution

The correct option is C 20 volt
$dV=−→E.→dr=−{Ex^i+Ey^j}.{dx^i+dy^j}$
$\int d V = - E x d x = - E_{0} x d x$ here,a=1
$Δ V = \int d v = \int_{0}^{4} - E_{0} x d x$
$= - \frac{E_{0} x^{2}}{2} = - 5 x^{2}$
=-5 (16)
V - 100 = -80
V =20 volt

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Electric field is space is given by →E=E0ax^j( E0=10N/C, a = 1 metre), find potential of point (4,3) (assuming potential at origin is 100 volt)

The correct option is C 20 voltdV=−→E.→dr=−{Ex^i+Ey^j}.{dx^i+dy^j} ∫dV=−Exdx=−E0xdx here,a=1 ΔV=∫dv=∫40−E0xdx =−E0x22=−5x2 =-5 (16) V - 100 = -80 V =20 volt

Electric field is space is given by $\to E = \frac{E_{0}}{a} x^j (E_{0} = 10 N / C$ , a = 1 metre), find potential of point (4,3) (assuming potential at origin is 100 volt)

The correct option is C 20 volt
$dV=−→E.→dr=−{Ex^i+Ey^j}.{dx^i+dy^j}$
$\int d V = - E x d x = - E_{0} x d x$ here,a=1
$Δ V = \int d v = \int_{0}^{4} - E_{0} x d x$
$= - \frac{E_{0} x^{2}}{2} = - 5 x^{2}$
=-5 (16)
V - 100 = -80
V =20 volt