Explanation
For equilateral triangle AOB, AO=OB=BA=a and AC=OC=BC=r
The potential at centroid C is V=k2qAC+k−qOC+k−qBC=k2qr−kqr−kqr=0
The net electric field at centroid C is E=k2qr2−kqr2cos30−kqr2cos30=k(0.2)2qr2≠0
As the total charge is zero so for dipole moment the choice of origin is independent. We assume O as origin.
dipole moment , p=−q(0)−q(Lˆi)+2q(L2ˆi+√3L2ˆj)
or p=√3qLˆj
Electric field due to charge sheet for charge density +σ is E=+σ2ϵ0 and for −σ is E=−σ2ϵ0
outside the plates field is E=+σ2ϵ0+−σ2ϵ0=0
between the plates field is E=σ2ϵ0+σ2ϵ0=σϵ0
Ans: (A),(C)
Hint: Find the net force on the dipole by using F=qE to calculate the force on both charges of the dipole.Explanation:Step 1: Reasons for incorrect options:
→τ=→P×→E
If the dipole moment P and electric field line E is in the same direction, in that case, torque will be zero in another case, it will have some non-zero value. So, option C is incorrect.
Answer:
Hence, option D is the correct answer.
Consider a Gaussian cynlinder of radius r and length l. Using Gauss's law the electric field at distance r is E.(2πrl)=λlϵ0⇒E=λ2πϵ0r
Here the electrostatic force is equal to centrifugal force. i.e,
qE=mv2r
qλ2πϵ0r=mv2r
orv2=qλ2πϵ0m
so v is independent of r.
The electric field at a distance r from the long wire is
E=14πϵ02λr=9×109×2×2.5×10−65×10−2=9×105NC−1
Hint: Find the net force on the dipole by using F=qE to calculate the force on both charges of the dipole.Explanation:Step 1: Concept used:Dipole consists of 2 equal and opposite charges +q and−q.
Consider a dipole, as shown below in the figure.
In uniform electric field E, if we place any dipole at that time,
Step 2: Calculation of force:Force on +q is qE and force on charge −q is −qE.
So, the total force will be sum of these two forces, and it will result in zero.
Hence, option A is the correct answer.
Consider a surface of a cylinder of radius r and length l as Gaussian surface. Apply Gauss's law, →E.→ds=Qenϵ0 where Qen is the charge enclosed by the Gaussian surface. here Qen=λl
thus, E.(2πrl)=λlϵ0
∴E=λ2πϵ0r=k2λr where k=14πϵ0r
The electric field due to the plate is E=σ2ϵ0
Here the force between the charged particle of charge q and the plate is
F=qE=qσ2ϵ0=2×10−6×4×10−62×8.854×10−12=0.45N
Using Gauss's law, the electric flux through the cube is ϕ=Qenϵ0 where Qen is the charge enclosed by cube.
As cube has six faces so the flux through each face is ϕe=ϕ6=Qen6ϵ0.
The flux is also the number of lines of force passing through the surface of cube. So
the number of lines of force through each face of the cube is n=ϕe=Qen6ϵ0=0.1×10−66×8.85×10−12∼1883
Hint: Like charges repel each other and unlike charges attract each other.
Part 1 : Figure for given situation
Consider that the soap is in negative charge,
\textbf{Part2: Explanation}
\bullet The charge will be distributed uniformly over the surface of the bubble by symmetry.
\bullet As we know, they want to step much farther apart, as charges repel, the only path outward, taking with them the soap surface.
\bullet At the same point, the rise in the restored force of the soap film (surface tension) would be equal and contrary to the electrostatic force, which will result in a new (larger) radius of equilibrium.
\bullet Because of the ionic similarities, this can happen to both positive and negatively charged bubbles.Therefore, when a negative charge is given to the soap bubble, then its radius will increase.
Hence option A correct.
The potential at a distance r due to dipole is \displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{\vec{p}.\hat{r}}{r^2} where \vec{p}= dipole moment.
or \displaystyle V=\dfrac{1}{4\pi\epsilon_0} \dfrac{p\cos\theta}{r^2}
or \displaystyle p=\dfrac{4\pi\epsilon_0 Vr^2}{\cos\theta}=\dfrac{4\pi\times 8.85\times 10^{-12}\times 2\times 10^{-5}\times (0.1)^2}{\cos30}=2.57 \times 10^{-17} Cm
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