Answer :
The electric field at x = 1.0 m is 12.18.
The electric field at x = 2.0 m is 0.135.
The electric potential at a point is a scalar quantity that is used to describe the electric potential energy per unit charge at that point. In the equation you provided, V is the electric potential at a point along the x-axis and x is the position along the x-axis. To find the electric field, which is denoted by Ex, you can use the equation Ex = -dV/dx, where dV is the change in electric potential and dx is the change in position along the x-axis.
To find the electric field at x = 1.0 m, you can plug in these values into the equation to get Ex = -dV/dx = -(-100e^(-2x))/dx = 50e^(-2x) = 50e^(-2*1.0) = 50e^(-2) = 12.18.
To find the electric field at x = 2.0 m, you can plug in these values into the equation to get Ex = -dV/dx = -(-100e^(-2x))/dx = 50e^(-2x) = 50e^(-2*2.0) = 50e^(-4) = 0.135.
In both cases, the electric field is positive, indicating that the electric field is directed away from the origin.
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