Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is
1 answer:
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
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• Electric current, can be said to be the flow of electric charge. These charges are through a conductor by moving electrons.
• Ammeter, is an instrument used to measure the amount of electric current in a circuit.
The name ammeter was derived from the unit of electric current (Amperes).
• Voltmeter, can be said to be an instrument used to measure the voltage(potential difference) in a circuit. It measures the electric potential between two points in an electrical circuit. It can also be called voltage meter.
• Electric circuit, can be defined as the conductive path for the flow of electric current.
It allows electric charge carriers to flow continuously.
• Resistance ,can be said to be the property of an electrical conductor which resists the flow of electric current. It is measured in ohms.
• Ohms law, states that the potential difference (V) between two points is directly proportional to the electric current across two points.
It is deonted as: V = I x R
Voltage = Current x Resistance
• Electric potential ,can be said to be the amount of electric potential energy at a point.
To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.
The expression for sound power is,
Here,
A = Area
I = Intensity
P = Power
At the same time the area can be written as,
Now the intensity is inversely proportional to the square of the distance from the source, then
The expression for the intensity at different distance is
Here,
= Intensity at distance 1
= Intensity at distance 2
= Distance 1 from light source
= Distance 2 from the light source
If we rearrange the expression to find the intensity at second position we have,
If we replace with our values at this equation we have,
Now using the equation to find the area we have that
Finally with the intensity and the area we can find the sound power, which is
Power is defined as the quantity of Energy per second, then