Electrostatics Concept Page - 4

Law

Vector form of coulomb's law

According to Coulomb's Law, electrostatic force on q1$q_1$ due to q2$q_2$ is given by:
F⃗ 12=14πεoq1q2r212r21^${\overrightarrow{F}}_{12}={\displaystyle \frac{1}{4\pi {\epsilon }_o}}{\displaystyle \frac{q_1{q}_2}{r_{12}^2}}\widehat{r_{21}}$
where
ϵo:${ϵ}_o:$ Permittivitty of the medium
q1$q_1$ and q2$q_2$: Magnitudes of Charges
r12:$r_{12}:$ Distance between charges
r21^:$\widehat{r_{21}}:$ Unit vector directed from q2$q_2$ to q1$q_1$

Example

SHM in electrostatic force

Example:
A particle of mass M$M$ and charge q$q$ is at rest at the mid point between two other fixed similar charges each of magnitude Q$Q$ placed a distance 2d$2d$ apart. The system is collinear as shown in figure. The particle is now displaced by a small amount x(x<<d)$x(x<<d)$ along the line joining the two charges and is left to itself. It will now oscillate about the mean position with what time period?(ε0=${\epsilon }_0=$ permittivity of free space)
Solution:Restoring force on displacement of x$x$
F=K[Qq(d−x)2−Qq(d+x)2]$F=K[\frac{Qq}{{(d-x)}^2}-\frac{Qq}{{(d+x)}^2}]$
=KQq[1(d−x)2−1(d+x)2]$=KQq[\frac{1}{{(d-x)}^2}-\frac{1}{{(d+x)}^2}]$
=KQq[4dx(d2−x2)2]$=KQq\left[\frac{4dx}{{(d^2-x^2)}^2}\right]$
=KQq[4dxd4](d>>x)$=KQq\left[\frac{4dx}{d^4}\right](d>>x)$
⇒F=KQq[4xd3]$\Rightarrow F=KQq\left[\frac{4x}{d^3}\right]$
Acceleration a=Fm=4KQqxMd3$a=\frac{F}{m}=\frac{4KQqx}{{Md}^3}$
or ω2=4KQqMd3${\omega }^2=\frac{4KQq}{{Md}^3}$
∴T=2πω=2πMd34KQq−−−−−−√$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{{Md}^3}{4KQq}}$
=2π3Md3ε0Qq−−−−−−−−√$=2\sqrt{\frac{{{\pi }^{3}Md}^3{\epsilon }_0}{Qq}}$

Definition

Superposition of electric charges

Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.This is termed as the principle of superposition.
Example : Consider three charges q1,q2,q3$q_1,q_2,q_3$ each equal to q at the vertices of an equilateral triangle of side l$l$. What is the force on a charge Q (with the same sign as q$q$) placed at the centroid of the triangle?
Solution In the given equilateral triangle ABC of sides of length l$l$, if we draw a perpendicular AD to the side BC,
AD=ACcos30=(3√/2)l$AD=ACcos30=(\sqrt{3}/2)l$ and the distance AO of the centroid O from A is (2/3)AD=(1/3√)l$(2/3)AD=(1/\sqrt{3})l$. By symmetry AO = BO = CO.
Force F1$F_1$ on Q due to charge q at A=3Qq4πϵ0l2$A={\displaystyle \frac{3Qq}{4\pi {ϵ}_0{l}^2}}$ along AOForce F2$F_2$ on Q due to charge q at B=3Qq4πϵ0l2$B={\displaystyle \frac{3Qq}{4\pi {ϵ}_0{l}^2}}$ along BO
Force F3$F_3$ on Q due to charge q at C=3Qq4πϵ0l2$C={\displaystyle \frac{3Qq}{4\pi {ϵ}_0{l}^2}}$ along CO
Therefore, the total force on Q=3Qq4πϵ0l2(r^−r^)=0$Q={\displaystyle \frac{3Qq}{4\pi {ϵ}_0{l}^2}}(\hat{r}-\hat{r})=0$ where r^$\hat{r}$ is the unit vector along OA.

Example

Method of image charges

Method of image charges is a very useful method which helps in finding force on a charge when placed near a conductor and charge is induced on it. An image charge of opposite nature is assumed at a point symmetrically opposite to the charge treating conductor as a mirror. Then, net force on the charge is given by the force between the image charge and the original charge removing the conductor from the picture. This is shown in the attached diagram. 
Net force is given by:
F=−q24πεo×4d2$F=-{\displaystyle \frac{q^2}{4\pi {\epsilon }_o\times 4d^2}}$
−$-$ sign shows that force is attractive.

Example

Electrostatic force along with other mechanical forces

Example:
Two similar helium filled balloons, each carrying charge Q, are tied to a 30 g weight and are floating at equilibrium as shown in the figure. Find Q.Solution:
From FBD of mass:
mg=Tcosθ+Tcosθ=2Tcosθ$mg=Tcos\theta +Tcos\theta =2Tcos\theta$..................................(1)
From FBD of balloon:
FE$F_E$:Electrostatic repulsive force
FE=14πε0Q2d2=14πε0Q236$F_E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{d^2}}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{36}}$
Tsinθ=FE$sin\theta =F_E$.........(2)
(2)÷(1),12tanθ=FEmg=38⇒FE=38mg$(2)÷(1),{\displaystyle \frac{1}{2}}tan\theta ={\displaystyle \frac{F_E}{mg}}={\displaystyle \frac{3}{8}}\Rightarrow F_E={\displaystyle \frac{3}{8}}mg$   (∵tanθ=34)$(\because tan\theta ={\displaystyle \frac{3}{4}})$
∴14πε0Q236=38×30×10−3×10$\therefore {\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{36}}={\displaystyle \frac{3}{8}}\times 30\times {10}^{-3}\times 10$
Q2=4.5×10−10$Q^2=4.5\times {10}^{-10}$
Q=21.2μC$Q=21.2\mu C$

Example

Force between two point charges

A charge q1$q_1$ exerts a force of 45N on a test charge q2=10−5C$q_2={10}^{-5}C$ located at a point 0.2m from q1$q_1$. The magnitude of q1$q_1$ is:We know F=kq1q2r2$F={\displaystyle \frac{k{q}_1{q}_2}{r^2}}$
⇒45=(9×109)(q1)(10−5)(0⋅2)2$\Rightarrow 45={\displaystyle \frac{(9\times {10}^9)(q_1)({10}^{-5})}{(0\cdot 2{)}^2}}$
⇒q1=45×0⋅2×0⋅29×109×10−5$\Rightarrow q_1={\displaystyle \frac{45\times 0\cdot 2\times 0\cdot 2}{9\times {10}^9\times {10}^{-5}}}$
⇒q1=2×10−5C$\Rightarrow q_1=2\times {10}^{-5}C$

Definition

Electric field lines

An electric field line is, in general, a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. An arrow on the curve is obviously necessary to specify the direction of electric field from the two possible directions indicated by a tangent to the curve. A field line is a space curve, i.e., a curve in three dimensions.

Definition

Electric Field due to point charges

These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges

Definition

Properties of electric field lines

The field lines follow some important general properties:
(i) Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.
(ii) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
(iii) Two field lines can never cross each other. (If they did,the field at the point of intersection will not have a unique direction, which is absurd.)
(iv) Electrostatic field lines do not form any closed loops.

Diagram

Electric field lines of simple configurations

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