Question

In one circuit, the terminals are at 98V and 100V. In the second circuit, they are at 2V and 6V. Why does the 2V and 6V produce a stronger electric field?

Answer 1

The second circuit with a voltage of 2V and 6V produces a stronger electric field than the first circuit with a voltage of 98V and 100V. The electric field strength is directly proportional to the voltage. A higher voltage will provide more force, resulting in a stronger electric field.

Step-by-step explanation:

The strength of an electric field is determined by the voltage or potential difference between two points. In the given circuits, the second circuit with a voltage of 2V and 6V produces a stronger electric field than the first circuit with a voltage of 98V and 100V.

The electric field strength is directly proportional to the voltage. The higher the voltage, the stronger the electric field. In the second circuit, the voltage difference is larger, resulting in a stronger electric field between the terminals.

For example, imagine the voltage as a pressure pushing the charges in a circuit. A higher voltage will provide more force, resulting in a stronger electric field.

Answer 2

The potential difference between circuit 1 = 100V - 98V

= 2V

The potential difference between circuit 2 = 6V - 2V

= 4V

According to Ohm's law,

ΔV = RI

where,

ΔV is the potential difference

R is the resistance

I is the current

We can see that potential difference is directly proportional to the current i.e., greater the potential difference, greater will be the current and thus more will be the electric field.

And as per the potential difference of the two circuits, circuit 2 has greater potential difference than the circuit 1.

So, the circuit 2 will have greater produce stronger electric field.