The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. What is the magnitude and direction of the electric field?
1 answer:
The potential energy at x = 8 m is -2000 V and at x = 2 m is 400 V. The magnitude and direction of the electric field will be 400 V/m directed parallel to the +x-axis
Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.
The electric potential energy of any given charge or system of changes is termed as the total work done by an external agent in bringing the charge or the system of charges from infinity to the present configuration without undergoing any acceleration.
The relation between electric field and electric potential can be generally expressed as – “Electric field is the negative space derivative of electric potential.”
Electric field = - d V / dx
-(-2000-400) =
2400 = E (8-2)
2400 V = E (6)
E = 400 V/m
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