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Class 10 Class 12

Define electric flux. Write its SI units. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason.

The total number of electric lines of force passing normally through that area. It is given by:

SI unit of electric flux is Nm² C^–1. As the balloon is blown up, the total charge on the balloon surface remains unchanged, so the total electric flux coming out of its surface remains unchanged.

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Define electric field intensity. Write its SI unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length 2a at the mid-point of the line joining the two charges.

The force experienced by a unit positive charge placed at that point is termed as the electric field intensity.

The SI unit of electric field intensity = NC^–1Electric field at any equatorial point of a dipole is
The force experienced by a unit positive charge placed at that point

At the mid-point of the dipole, r = 0,
Hence,

straight E with rightwards arrow on top space equals space minus fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight P with rightwards arrow on top over denominator straight a cubed end fraction

The direction of E is from positive charge to negative charge.

2722 Views

Calculate the Coulomb force between 2 alpha particles separated by 3.2 x 10^–15 m.

Given,
Charge on an alpha particle, q₁ = q₂ = +2e
Distance between the particles, r = 3.2 x 10^–15 m

Now, using coulomb's law, we get
Force acting on the particles as,

F = \frac{1}{4 {πε}_{o}} \frac{q_{1} q_{2}}{r^{2}} F = \frac{9 \times 10^{9} \times 2 \times 1.6 \times 10^{- 19} \times 2 \times 1.6 \times 10^{- 19}}{3.2 \times 10^{- 15} \times 3.2 \times 10^{- 15}} N = 90 N

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The electric field E due to a point charge at any point near it is defined as $E = \lim_{q \to 0} \frac{F}{q},$ where q is the test charge and F is the force acting on it. What is the physical significance of $\lim_{q \to 0}$ in this expression? Draw the electric field lines of a point charge Q when (i) Q > 0 and (ii) Q < 0

It is indicated by the

limit as straight q rightwards arrow 0 of

that the test charge q is so small that its presence does not disturb the distribution of source charge and therefore, its electric field. The electric fields of the point charge Q are shown below.
It is indicated by the that the test charge q is so small that i

It is indicated by the that the test charge q is so small that i

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A copper sphere of mass 2 g contains nearly 2 x 10²² atoms. The charge on the nucleus of each atom is 29 e. What fraction of the electrons must be removed from the sphere to give it a charge of +2 μC?

Total number of electrons in the sphere = 29 x 2 x 10^{22
$No . of electrons = \frac{q}{e} = \frac{2 \times 10^{- 6}}{1.6 \times 10^{- 19}} = 1.25 \times 10^{13}$
$Fraction of electrons removed = \frac{1.25 \times 10^{13}}{29 \times 2 \times 10^{22}} = 2.16 \times 10^{- 11}$}

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Electric Charges and Fields

The electric field E due to a point charge at any point near it is defined as E = limq→0Fq, where q is the test charge and F is the force acting on it. What is the physical significance of limq→0 in this expression? Draw the electric field lines of a point charge Q when (i) Q > 0 and (ii) Q < 0