Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrostatics question

2021 · 25 Feb · Shift 1 · Q69
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Electrostatics
  5. /2021 · 25 Feb · Shift 1 · Q69

Electrostatics question

2021 · 25 Feb · Shift 1 · Q69

JEE MainPhysicsElectrostaticsNumerical+4 / −1
The electric field in a region is given by E→=(35E0i^+45E0j^)NC\overrightarrow E = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right){N \over C}E=(53​E0​i+54​E0​j​)CN​. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y −-− z plane) to that of the surface of area 0.3 m2 (parallel to x −-− z plane) is a : b, where a = ‾\underline{\hspace{2cm}}​ [Here i^{\widehat i}i, j^{\widehat j}j​ and k^{\widehat k}k are unit vectors along x, y and z-axes respectively.]
Numerical answer
View written solutionFree

Correct answer: 1

  1. The electric field is
E⃗=(35E0i^+45E0j^)\vec E = \left(\frac{3}{5}E_0\hat i + \frac{4}{5}E_0\hat j\right)E=(53​E0​i^+54​E0​j^​)

So its components are:

Ex=35E0,Ey=45E0E_x = \frac{3}{5}E_0, \qquad E_y = \frac{4}{5}E_0Ex​=53​E0​,Ey​=54​E0​
  1. Flux through a surface is
Φ=E⃗⋅A⃗\Phi = \vec E \cdot \vec AΦ=E⋅A

where A⃗\vec AA is the area vector normal to the surface.

  1. For the rectangular surface parallel to the yzyzyz-plane:
  • its area vector is along ±i^\pm \hat i±i^
  • area = 0.2 m20.2\,\text{m}^20.2m2

Hence flux magnitude through this surface is

Φ1=Ex×0.2=35E0×0.2\Phi_1 = E_x \times 0.2 = \frac{3}{5}E_0 \times 0.2Φ1​=Ex​×0.2=53​E0​×0.2 Φ1=325E0\Phi_1 = \frac{3}{25}E_0Φ1​=253​E0​
  1. For the rectangular surface parallel to the xzxzxz-plane:
  • its area vector is along ±j^\pm \hat j±j^​
  • area = 0.3 m20.3\,\text{m}^20.3m2

Hence flux magnitude through this surface is

Φ2=Ey×0.3=45E0×0.3\Phi_2 = E_y \times 0.3 = \frac{4}{5}E_0 \times 0.3Φ2​=Ey​×0.3=54​E0​×0.3 Φ2=625E0\Phi_2 = \frac{6}{25}E_0Φ2​=256​E0​
  1. Therefore,
Φ1:Φ2=325E0:625E0=1:2\Phi_1 : \Phi_2 = \frac{3}{25}E_0 : \frac{6}{25}E_0 = 1:2Φ1​:Φ2​=253​E0​:256​E0​=1:2

So in the ratio a:ba:ba:b, we get

a=1a = 1a=1
  1. Comparison with stored answer:
  • Derived answer: 111
  • Stored correct answer: 111

They match.

PreviousNext

More from Electrostatics

  • 512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ​ V.2021 · Numerical
  • A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is : Includes diagram2021 · MCQ
  • Two identical conducting spheres with negligible volume have 2.1 nC and − 0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is ​×…2021 · Numerical
  • Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is 21a​×10−8 C. The…2021 · Numerical
  • A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through…2021 · Numerical
  • Two ideal electric dipoles A and B, having their dipole moment p1 and p2 respectively are placed on a plane with their centres at O as shown in the figure. At point C on the axis of dipole A, the resultant electric field is making an angle… Includes diagram2021 · MCQ
  • A solid metal sphere of radius R having charge q is enclosed inside the concentric spherical shell of inner radius a and outer radius b as shown in the figure. The approximate variation electric field E as a function of… Includes diagram2021 · MCQ
  • The two thin coaxial rings, each of radius 'a' and having charges +Q and − Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is :2021 · MCQ