Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and
whose direction of fire can be adjusted necessarily and sufficiently fast, is (t) = 30 x 10m +20 x 10'my. Besides that, the position of a target is determined as Fr(t) =(-12 x 10 m + 100). Thus at which instant of time must the artillery be fired in order that the target is terminated as earlier as possible? (Ignore the friction, the air resistance and assume that there is no Coriolis force)
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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
To learn more about electric potential energy here
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