<em>Answer: D</em>
<em>Given data:</em>
The electric field due to charge q₁ is (E₁) = 1.5 × 10⁵ N/C
The electric field due to charge q₂ is (E₂) = 7.2 × 10⁵ N/C
Determine the net electric field at point P (Enet) = ?
electric field measured in Newtons/coulomb
<em>We know that,</em>
The net electric field (Enet) is vector sum of E₁ and E₂ caused by the charge q₁ and q₂ respectively
Enet = E₁ + E₂
= (1.5 × 10⁵)+ (7.2 × 10⁵)
= 8.7 × 10⁵ Newtons/coulomb
<em>The net electric field at point P is 8.7 × 10⁵ N/C</em>
Answer:
The mass of the meter stick is 
Explanation:
Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero
Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at 
Now lets consider the later case where stick is balanced by two 5.12 g coins .
Here torque due to two coins = 
Torque due to weight of meter stick =
= 
where m = mass of the meter stick
Here
.
Upon equating we will be getting mass of the meter stick =
Hello!
We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.
Refer to the free-body diagram in the image below for clarification.
We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)
The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.
Therefore:

We can solve the equation for 'v':

Plug in values and solve.

E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V
You should charge it to 1000 volts to store 1.0 J of energy.