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Preeti Dabral 2 years, 11 months ago
Self-efficacy refers to an individual's belief in his or her capacity to execute behaviors necessary to produce specific performance attainments (Bandura, 1977, 1986, 1997). Self-efficacy reflects confidence in the ability to exert control over one's own motivation, behavior, and social environment.
Aarushi Parihar 2 years, 10 months ago
Anjan Karthi 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months 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 2 years, 11 months ago
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🤟Royal Thakur 🤟 2 years, 11 months ago
Aseem Mahajan 2 years, 11 months ago
Posted by Aayush Singh 2 years, 11 months ago
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Preeti Dabral 2 years, 11 months 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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Navneet Kaur 2 years, 11 months ago
1Thank You