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il63 [147K]
3 years ago
8

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

here? Explain with graph.​
Physics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

(i)  Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

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A 5 Kg mass is suspended from a spring. The spring is stretched 8 cm from equilibrium. What is the spring constant? Use g=9.8 m/
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The spring constant is:

                                             \Large\displaystyle\text{$\begin{aligned}k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned}$}

To calculate the <em>spring constant</em> we must remember the law for it, the Hooke's Law:

                                            \Large\displaystyle\text{$\begin{aligned}\vec{F} = -k\Delta \vec{x} \end{aligned}$}

Where k is the spring constant [N/m].

So if the mass is suspended it means that its weight is equal to the elastic force (values), then we can write:

                                           \Large\displaystyle\text{$\begin{aligned}\vec{W} = k\Delta \vec{x} \end{aligned}$}

Therefore:

                                          \Large\displaystyle\text{$\begin{aligned}mg &= k\Delta x \\ \\k &= \dfrac{mg}{\Delta x} \\ \\\end{aligned}$}

Now we just have to put the values and calculate:

                                              \Large\displaystyle\text{$\begin{aligned}k &= \dfrac{mg}{\Delta x} \\ \\k &= \dfrac{5\cdot 9.8}{0.08} \\ \\k &= 612.5\ \dfrac{\text{N}}{\text{m}}\end{aligned}$}

I hope you liked it

Any doubt? write in the comments and I'll help you

7 0
3 years ago
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