Question

pls do 3 and 7 ques please give proper explanation

Answer 1

check the solution of 3rd question and follow me

Answer 2

(3).

Answer:(3). The number of additional electron will be $n = 15\times10^{14}$.

Explanation:

Given that,

Charge q = 80 micro coulomb

We know that,

charge of electron $e = 1.6\times10^{-19}\ C$

$Q = ne$

$n = \dfrac{Q}{e}$

$n = \dfrac{80\times10^{-6}}{1.6\times10^{-19}}$

$n = 15\times10^{14}$

Hence, The number of additional electron will be $n = 15\times10^{14}$.

#7. Answer: (b).

The acceleration of central charge is 9 m/s²(downward)

Explanation:

Given that,

Magnitude of the each charge $q= 10^{-5}\ C$

Mass of the each charge = 1 kg

Magnitude of the central charge $q= 5\times10^{-5}\ C$

Mass of the central charge = 0.5 kg

The distance of the each charge from the center charge d = 1 m

We know that,

The force on charge $P_{2}$ is

$F_{2} = \dfrac{1}{4\pi \epsilon_{0}}\dfrac{qq_{2}}{r^{2}}$

$F_{2} = 9\times10^{9}\times\dfrac{10^{-5}\times5\times10^{-5}}{1}$

$F_{2} = 4.5 N$

Similarly,

$F_{3}=F_{4}= F_{5}$

So, $F_{2}=F_{3}=F_{4}= F_{5} =F = 4.5 N$

The resultant horizontal force is zero.

The resultant vertical force is

$F_{R}= 2\times4.5(cos36-cos72)$

$F_{R}= 4.5 N$

The acceleration is

$a = \dfrac{F}{m}$

$a = \dfrac{4.5}{.5}=9\text{ m/s}^2$

Hence, acceleration of central charge is 9 m/s²(downward).