#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Hello,This is example from Physics NCERT Book 12th ! and i want to know how this Electric field is calulated for this both left and right faces !Here is QuestionExample 1.11 The electric field components in Fig. 1.27 areEx = áx1/2, Ey = Ez = 0, in which á = 800 N/C m1/2. Calculate (a) theflux through the cube, and (b) the charge within the cube. Assumethat a = 0.1 m.  Sol. in NCERT :Since the electric field has only an x component, for facesperpendicular to x direction, the angle between E and ΔS is± π/2. Therefore, the flux φ = E.ΔS is separately zero for each faceof the cube except the two shaded ones. Now the magnitude ofthe electric field at the left face isEL = αx1/2 = αa1/2(x = a at the left face).The magnitude of electric field at the right face isER = α x1/2 = α (2a)1/2(x = 2a at the right face).The corresponding fluxes areφL= EL.ΔS = ΔS EL ⋅ nˆ L =EL ΔS cosθ = –EL ΔS, since θ = 180°= –ELa2φR= ER.ΔS = ER ΔS cosθ = ER ΔS, since θ = 0°= ERa2Net flux through the cube= φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2]= αa5/2 ( 2 –1)= 800 (0.1)5/2 ( 2 –1)= 1.05 N m2 C–1(b) We can use Gauss’s law to find the total charge q inside the cube.We have φ = q/ε0 or q = φε0. Therefore,q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C.

arun
123 Points
5 years ago
it is given that EX = $\alpha$x^(1/2) in this ‘x’ represents the distance from origin on x-axis. In figure you can see left face is at a distance of left face from origin is ‘a’ so,
EX (for left face) = $\alpha \sqrt{a}$
and right face is at a distane of ‘2a’ so,
EX (for right face) =$\alpha \sqrt{2a}$
Rishi Sharma
one year ago
Hello Mark,
Let us understand the problem in this way,
Electric Flux is defined as a number of electric field lines, passing per unit area.
It is also defined as the product of electric field and surface area projected in a direction perpendicular to the electric field.
As per given data electric field is in the x axis direction thats why we need to consider both the face perpendicular to x axis rest all are parallel to x axis. Hence no flux will develop.
Thanks
I hope above solution will clear your all doubts.
Please feel free to post and ask as much doubts as possible.
All the best.