1If the distance between two point charges is halved, the electric intensity becomes:aHalfb1/4 timescDoubled4 timesShow answer
3Presence of dielectric always:aIncrease the electrostatic forcebDecrease the electrostatic forcecDoes not effectsdDouble the electrostatic forceShow answer
6In SI units the value of permittivity of free space ε₀ is:a9×109Nm2c−2b8.85×10−12Nm2C−2c8.85×10−12N−1m−2C−2d9×109N−1m2C2Show answer
7The electrostatic force of repulsion between two electrons at a distance 1m is:a2.3×10−24Nb2.3×10−26Nc2.3×10−28Nd2.3×10−30NShow answer
9For computation of electric flux, the surface area should be:aParallelbFlatcCurveddSphericalShow answer
10The force between two-point charges separated by air in 4N when separated by a medium of relative permittivity 2, the force between them become:a1/2Nb2Nc4Nd8NShow answer
12If the distance between the two charges bodies is halved, the force between them becomes:aDoublebHalfcFour timesdOne fourthShow answer
13The equation small -Δvvr is called:aElectric potentialbElectric energycPotential barrierdPotential gradientShow answer
14If the medium between the charges is not free space then electrostatic force will:aIncreasebDecreasecRemain samedNone of theseShow answer
15For which material medium, force between two charges particles is maximum?aAmmoniabGermaniumcMicadTeflonShow answer
16The negative of the potential gradient is:aElectrostatic forcebElectromotive forcecPotential differencedElectric field intensityShow answer
18Closeness of the electric field lines is the measure of:aDirection of fieldbStrength of fieldcPotential differencedUniformity of fieldShow answer
19The number of electrons in one coulomb charge is equal to:a1.6×10−19b6.25×10−19c6.25×1018d6.25×1019Show answer
20Identify the practical application of electrostatic force:aInkjet printerbX-rayscLaserdA.C generatorShow answer
