Hi there!
Recall the equation for electric potential of a point charge:

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)
Q = Charge (C)
r = distance (m)
We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.
Upper right charge's potential:

Lower left charge's potential:

Add the two, and subtract from the total EP at the point:

The remaining charge must have a potential of 2036.25 V, so:

Answer:
Minimum diameter of the camera lens is 22.4 cm
The focal length of the camera's lens is 300cm
Explanation:
y = Resolve distance = 0.3 m
h = Height of satellite = 100 km
λ = Wavelength = 550 nm
Angular resolution

From Rayleigh criteria

Minimum diameter of the camera lens is 22.4 cm
Relation between resolvable feature, focal length and angular resolution

The focal length of the camera's lens is 300cm
Answer:
<h2><em>
6000 counts per second</em></h2>
Explanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
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<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>
Maybe the word could be converted?