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Electrostatics Mock Tests

214 questions available

Electrostatics Mock Test 1

Questions: 30

Electrostatics Mock Test 2

Questions: 30

Electrostatics Mock Test 3

Questions: 30

Electrostatics Mock Test 4

Questions: 30

Electrostatics Mock Test 5

Questions: 30

Electrostatics Mock Test 6

Questions: 30

Electrostatics Mock Test 7

Questions: 30

Electrostatics Mock Test 8

Questions: 4

Sample Questions

NEET Physics
A spherical conductor of radius 10 cm has a charge of 3.2 × 10⁻⁶ C distributed uniformly. The electric field at a point 5 cm from the centre is:
A Zero
B 2.88 × 10⁶ N/C
C 5.76 × 10⁶ N/C
D 1.44 × 10⁶ N/C
BITSAT Physics
Two point charges q₁ = 2 × 10⁻⁶ C and q₂ = 3 × 10⁻⁶ C are separated by a distance of 0.1 m in vacuum. What is the electrostatic force between them? (Take k = 9 × 10⁹ N·m²/C²)
A 5.4 N
B 2.7 N
C 10.8 N
D 1.8 N
JEE Advanced Physics
A solid non-conducting sphere of radius R has charge density ρ(r) = ρ₀·(1 - r/R). The electric field at distance r < R from centre is proportional to:
A r·(4R - 3r)
B R - r
C r²·(R - r)
D R² - r²
BITSAT Physics
A point charge q is at the center of a cube. The electric flux through one face of the cube is:
A q/ε₀
B q/6ε₀
C q/4πε₀
D q/12ε₀
EAPCET Physics
An electric dipole with dipole moment p⃗ = pî is placed in an electric field E⃗ = E₀(1 + x/L)î where E₀ and L are constants. The magnitude of the net force and torque on the dipole are respectively:
A (pE₀/L) and 0
B (pE₀/L) and pE₀
C 0 and pE₀
D (2pE₀/L) and 0
BITSAT Physics
A parallel plate capacitor with air between plates has capacitance C. A dielectric slab of constant K and thickness t = 2d/3 (where d is plate separation) is inserted. The new capacitance is:
A 3CK/(K + 1)
B 3CK/(K + 2)
C 3CK/(2K + 1)
D CK/(K + 1)
VITEEE Physics
A parallel plate capacitor has capacitance C in air. If a dielectric of constant K = 4 is inserted between the plates, the new capacitance will be:
A 4C
B C/4
C C
D 2C
BITSAT Physics
A parallel plate capacitor with air between plates has capacitance C0. If a dielectric slab of constant K and thickness equal to half the plate separation is inserted, the new capacitance is:
A 2KC0/(K+1)
B KC0
C 2C0/(K+1)
D C0(K+1)/2

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