Answer:
12500 V
Explanation:
The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:
where
is the potential difference between the plates
E is the electric field strength
d is the distance between the plates
For the capacitor in this problem, we have
Substituting, we find
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
The electric field is zero at x = -16.45cm
Data;
As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.
Let the point be at x.
For an electric field to be equal to zero;
Let's solve for x using mathematical methods.
Solving the above quadratic equation;
The electric field is zero at x = -16.45cm
Learn more on electric field at a point here;