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

2022 · 28 Jun · Shift 2 · Q59
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. /2022 · 28 Jun · Shift 2 · Q59

Electrostatics question

2022 · 28 Jun · Shift 2 · Q59

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two point charges A and B of magnitude +8 ×\times× 10 −-− 6 C and −-− 8 ×\times× 10 −-− 6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 ×\times× 104 NC −-− 1. The distance 'd' between the point charges A and B is :
  1. A
    2.0 m
  2. B
    3.0 m
  3. C
    1.0 m
  4. D
    4.0 m
View written solutionFree

Correct answer: B

  1. Given data
  • Charge at AAA: +8×10−6 C+8 \times 10^{-6}\,\text{C}+8×10−6C
  • Charge at BBB: −8×10−6 C-8 \times 10^{-6}\,\text{C}−8×10−6C
  • Distance between charges: ddd
  • Midpoint between them: OOO
  • Electric field at midpoint: 6.4×104 N C−16.4 \times 10^4\,\text{N C}^{-1}6.4×104N C−1
  1. Electric field at the midpoint due to each charge

The midpoint is at a distance

d2\frac{d}{2}2d​

from each charge.

Electric field due to one point charge is

E=kqr2E = \frac{kq}{r^2}E=r2kq​

where k=9×109 N m2/C2k = 9 \times 10^9\,\text{N m}^2\text{/C}^2k=9×109N m2/C2.

So, field at OOO due to charge AAA is

EA=k(8×10−6)(d/2)2E_A = \frac{k(8 \times 10^{-6})}{(d/2)^2}EA​=(d/2)2k(8×10−6)​

Similarly, field at OOO due to charge BBB is

EB=k(8×10−6)(d/2)2E_B = \frac{k(8 \times 10^{-6})}{(d/2)^2}EB​=(d/2)2k(8×10−6)​
  1. Direction of electric fields
  • Due to positive charge AAA, field at OOO is away from AAA.
  • Due to negative charge BBB, field at OOO is towards BBB.

At the midpoint, both these directions are the same, so the fields add.

Hence total field is

E=EA+EB=2⋅k(8×10−6)(d/2)2E = E_A + E_B = 2\cdot \frac{k(8 \times 10^{-6})}{(d/2)^2}E=EA​+EB​=2⋅(d/2)2k(8×10−6)​
  1. Substitute the given electric field
6.4×104=2⋅9×109⋅8×10−6(d/2)26.4 \times 10^4 = 2\cdot \frac{9 \times 10^9 \cdot 8 \times 10^{-6}}{(d/2)^2}6.4×104=2⋅(d/2)29×109⋅8×10−6​

Now,

9×109⋅8×10−6=72×1039 \times 10^9 \cdot 8 \times 10^{-6} = 72 \times 10^39×109⋅8×10−6=72×103

So,

6.4×104=2⋅72×103(d/2)26.4 \times 10^4 = 2\cdot \frac{72 \times 10^3}{(d/2)^2}6.4×104=2⋅(d/2)272×103​ 6.4×104=144×103(d/2)26.4 \times 10^4 = \frac{144 \times 10^3}{(d/2)^2}6.4×104=(d/2)2144×103​

Thus,

(d/2)2=144×1036.4×104(d/2)^2 = \frac{144 \times 10^3}{6.4 \times 10^4}(d/2)2=6.4×104144×103​ (d/2)2=1446.4×10−1(d/2)^2 = \frac{144}{6.4} \times 10^{-1}(d/2)2=6.4144​×10−1 (d/2)2=22.5×10−1=2.25(d/2)^2 = 22.5 \times 10^{-1} = 2.25(d/2)2=22.5×10−1=2.25

Therefore,

d2=1.5\frac{d}{2} = 1.52d​=1.5

So,

d=3.0 md = 3.0\,\text{m}d=3.0m
  1. Option check
  • A: 2.0 m2.0\,\text{m}2.0m ❌
  • B: 3.0 m3.0\,\text{m}3.0m ✅
  • C: 1.0 m1.0\,\text{m}1.0m ❌
  • D: 4.0 m4.0\,\text{m}4.0m ❌

Therefore, the correct answer is B.

PreviousNext

More from Electrostatics

  • A spherically symmetric charge distribution is considered with charge density varying as ρ(r)={ρ0​(43​−Rr​) zero ​ for r≤R for r>R​… Includes diagram2022 · MCQ
  • Given below are two statements. Statement I : Electric potential is constant within and at the surface of each conductor. Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at…2022 · MCQ
  • Two identical metallic spheres A and B when placed at certain distance in air repel each other with a force of F. Another identical uncharged sphere C is first placed in contact with A…2022 · MCQ
  • A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC − 1. If the charge on the particle is 40 μ C and the initial velocity is 200 ms − 1, how much distance…2022 · MCQ
  • Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :2022 · MCQ
  • If the electric potential at any point (x, y, z) m in space is given by V = 3x2 volt. The electric field at the point (1, 0, 3) m will be :2022 · MCQ
  • Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ​ E. Where E is the potential energy of a single…2022 · Numerical
  • A cube is placed inside an electric field, E=150y2j​. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is : Includes diagram2021 · MCQ