An insulating sphere of radius 13 cm has a uniform charge density throughout its volume. 13 cm 21.6 cm 7.5 cm p If the magnitude of the electric field at a distance of 7.5 cm from the center is 78400 N/C , what is the magnitude of the electric field at 21.6 cm from the center? Answer in units of N/C
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Ответ:
The value is![E_1 = 49224.1 \ N/C](/tpl/images/0771/9828/fecf6.png)
Explanation:
From the question we are told that
The radius is![r = 13 \ cm = 0.13 \ m](/tpl/images/0771/9828/985d9.png)
The electric field is
at a distance ![d = 7.5 \ cm = 0.075 \ m](/tpl/images/0771/9828/b75ae.png)
Generally the electric field at a distance
is mathematically represented as
formula is because![d < r](/tpl/images/0771/9828/131b1.png)
Here k is the coulomb constant with value![k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.](/tpl/images/0771/9828/e34bb.png)
=>![78400= \frac{9*10^{9} * q * 0.075}{0.13^3 }](/tpl/images/0771/9828/8a510.png)
Generally the electric field at a distance
is mathematically represented as
formula is because![d r](/tpl/images/0771/9828/cc7ef.png)
Now dividing![E_1\ \ by \ \ E](/tpl/images/0771/9828/01c21.png)
=>![E_1 = \frac{0.13 ^3}{ 0.075 * 0.216^2} * 78400](/tpl/images/0771/9828/9c632.png)
=>![E_1 = 49224.1 \ J](/tpl/images/0771/9828/d69e3.png)
Ответ:
after 9second
displacement = 9*5 = 45m