Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N
2) Charge
Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]
k = 9.00 * 10^9 N*m^2 / C^2
charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]
charge = 0.0000001 C = 0.0001 mili C
Answer:
Exercise 1;
The centripetal acceleration is approximately 94.52 m/s²
Explanation:
1) The given parameters are;
The diameter of the circle = 8 cm = 0.08 m
The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m
The speed of motion = 7 km/h = 1.944444 m/s
The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²
The centripetal acceleration ≈ 94.52 m/s²
Answer:
it will be 1/√2 of its original period.
Explanation:
Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E = 
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero