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Ask QuestionPosted by Ajitha Kumari 3 years, 1 month ago
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Anjan Karthi 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
- The electric flux through an area is defined as the electric field multiplied by the area of the surface projected on a plane, perpendicular to the field. Its S.I. unit is voltmeter (Vm) or Newton metre square per coulomb (Nm2 C-1). The given statement is justified because while measuring the flux, the surface area is more important than its volume on its size.
- Electric field inside the shell:
The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let's consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward.
Electric flux through the Gaussian surface is given by,
{tex}=\int_{s} \vec{E}_{i} \cdot d \vec{S}{/tex}
{tex}=\int E_{i} d S \cos 0=E_{i} .4 \pi r^{2}{/tex}
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Therefore, using Gauss's theorem, we have
{tex}\int_{S} \vec{E}_{i} \cdot d \vec{S}=\frac{1}{\epsilon_{0}} \times \text { charge enclosed }{/tex}
{tex}\Rightarrow E_{i} \cdot 4 \pi r^{2}=\frac{1}{\epsilon_{0}} \times 0{/tex}
{tex}\Rightarrow{/tex} Ei = 0
Thus, electric field at each point inside a charged thin spherical shell is zero.
Mayank Chauhan 3 years, 1 month ago
Posted by Aayush Singh 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
r=√(2mqV) / qB
as given alpha particle q=2e=2×1.6×10^-19
V=10^4V
m=6.4×10^-27kg
B=2×10^-3T
r=√(2 × 6.4×10^-27 × 2× 1.6×10^-19 × 10^4) /1.6×10^-19 × 2×10^-3
=√(4096×10^-44) / 64 × 10^-23
=(64 × 10^-22) / (64 × 10^-23)
=10
therefore radius is 10m
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Kashu Verma 3 years, 1 month ago
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Anjan Karthi 3 years, 1 month ago
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Nidhi Rathod 3 years, 1 month ago
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Anjan Karthi 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
electron charge, (symbol e), fundamental physical constant expressing the naturally occurring unit of electric charge, equal to 1.602176634 × 10−19 coulomb.
Posted by Harleen Kaur 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
Based on the circuit given in the figure , we have three known resistance's R1,R2,R3 and an unknown variable resistor RX, a source of voltage, and a sensitive ammeter.
Kirchhoff's first rule is applied to find the currents in junctions B and D:
I3−Ix+Ig=0
I1−I2−Ig=0
Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:
I3.R3−Ig.Rg−I1.R1=0
Ix.RX−I2.R2+Ig.Rg=0
The bridge is balanced and Ig = 0, so the second set of equations can be rewritten as:
I3.R3=I1.R1
Ix.RX=I2.R2
From first rule of Kirchoff.
I3=Ix
I1=I2
{tex}\begin{aligned} & \frac{\mathrm{I}_3 \cdot \mathrm{R}_3}{\mathrm{I}_{\mathrm{n}} \cdot \mathrm{R}_{\mathrm{X}}}=\frac{\mathrm{I}_1 \cdot \mathrm{R}_1}{\mathrm{I}_2 \cdot \mathrm{R}_2} \\ & \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_{\mathrm{x}}} \end{aligned}{/tex}
Posted by Rachna Yadav 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. Typically, the reference point is Earth, although any point beyond the influence of the electric field charge can be used.
Yash Thakkar 3 years, 1 month ago
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Anjan Karthi 3 years, 1 month ago
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Preeti Dabral 3 years, 1 month ago
A half-wave rectifier converts an AC signal to DC by passing either the negative or positive half-cycle of the waveform and blocking the other. Half-wave rectifiers can be easily constructed using only one diode, but are less efficient than full-wave rectifiers.
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Anjan Karthi 3 years, 1 month ago
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Preeti Dabral 3 years, 2 months ago
12 cm
As the charge q, can be equilibrium present only when its is kept in between charges +9e and +e, correct answer is x=12 cm.
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Anjan Karthi 3 years, 1 month ago
1Thank You