Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)
Answer: v = 5.79 * 10^10m/s.
Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.
Mathematically, we have that
qV = 1/2mv²
q= magnitude of an electronic charge = 1.609*10^-16c
V= potential difference = 95v
m = mass of an electronic charge = 9.11* 10^-31kg.
v = velocity of electron.
Let us substitute the parameters, we have that
1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2
1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²
305.71 * 10^-16 = 9.11 * 10^-31 * v²
v² = 305.71 * 10^-16/ 9.11 * 10^-31
v² = 3.355 * 10^21.
v = √3.355 * 10^21
v = 5.79 * 10^10 m/s
A=(Vo-V)/t=(10-0)/0.26=10/0.26=38.46m/s2
F=ma=70*38.46=2692.3N
The strength of the electric field is generally equal to
E = F/Q
However, this could be modified in terms of charge and distance,d.
E = kQ/d²
The k is actually a constant which is equal to 1/4π∈₀
Since no given quantitative data, the answer should just be E = kQ/r².
Answer:
<em>The kinetic energy increases by a factor of 9
</em>
Explanation:
<u>Kinetic Energy
</u>
It's the energy related to the speed of an object and is calculated by the formula
Let's suppose the initial speed of the mass is 
, then the kinetic energy is
Now the speed will triple, i.e. 
. The new kinetic energy is
Thus
The kinetic energy increases by a factor of 9
Let's check with numbers: 
The relation is
The new kinetic energy is nine times the initial kinetic energy.
