Question

please answer this question​

Answer 1

option C is the correct answer

Qi= epsilon ( E1.A- E2.A2)

where E1 is electric field at 2nd surface which is at a distance 2m from origin

and E2 is electric field at 1st surface which is at a distance 1m from origin

Answer 2

QUESTION:

The electric field components in space is given as $E_x = \alpha x^{1/3}, \ E_y = 0, \ E_z = 0 \ where \ \alpha = 400 N/C m^{1/3}$ ans x is the distance from the origin. The charge enclosed within the cube as shown in the figure is

ANSWER:

• $q_{en} = 8.85 \times {10}^{ - 10}C$

GIVEN:

• $E_x = \alpha x^{1/3}, \ E_y = 0, \ E_z = 0 \ where \ \alpha = 400 N/C m^{1/3}$

• a = 1 m

TO FIND:

• The charge enclosed within the cube.

EXPLANATION:

$\boxed{ \bold{ \large{ \phi = Eds \cos \theta}}}$

• $E_L$ = Electric field in the left side of the cube.

• $E_R$ = Electric field in the right side of the cube.

• $\phi_L$ = Electric flux in the left side of the cube.

• $\phi_R$ = Electrc flux in the right side of the cube.

$\phi_L = E_Lds \cos \theta$

$E_L= \alpha a^{1/3}\\ \\ \\ E_L= 400 \times 1^{1/3}\\ \\ \\E_L= 400 N/C \\ \\ \\ Area \ of \ cube = a^3 \\ \\ \\ds = 1^3 = 1 \\ \\ \\ \theta = 180^{\circ}\ as \ ds \ and \ E_L \ are \ in \ opposite \\ direction\\ \\ \\ \phi_L = 400(1) \cos 180^{ \circ} \\ \\ \\ \phi_L = - 400Nm^2/C$

$\phi_R = E_Rds \cos \theta$

$E_R= \alpha (2a)^{1/3}\\ \\ \\ E_R= 400\times 2^{1/3}\\ \\ \\E_R= 400\times 2^{1/3} N/C \\ \\ \\ Area \ of \ cube = a^3 \\ \\ \\ds = 1^3 = 1 \\ \\ \\ \theta = 0^{\circ}\ as \ ds \ and \ E_R \ are \ in \ same \\ direction\\ \\ \\ \phi_R= 400(2^{1/3}) \cos 0^{ \circ} \\ \\ \\ \phi_R=400(2^{1/3})Nm^2/C$

$\phi_{net} = E_R +E_L$

$\phi_{net} = 400 \times {2}^{ \frac{1}{3} } -400 \\ \\ \\ \phi_{net} = 400( {2}^{ \frac{1}{3} } -1) \\ \\ \\ \phi_{net} =400(1.25 - 1) \\ \\ \\ \phi_{net} =400(0.25) \\ \\ \\ \phi_{net} =100 Nm^2/C$

$\boxed{ \bold{ \huge{ \phi = \dfrac{q_{en}}{\epsilon_o}}}}$

$\boxed{ \bold{ \huge{ \phi\epsilon_o= q_{en}}}}$

$\phi_{net}\epsilon_o = 100 \times 8.85 \times {10}^{ - 12} \\ \\ \\ q_{en} = 8.85 \times {10}^{ - 10}C$

NOTE: REFER ATTACHMENT FOR DIAGRAM.