The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.6 V and b = -4.90 V/m. (a) Determine the potential at x = 0. 10.6 V Determine the potential at x = 3.00 m. V (b) Determine the magnitude and direction of the electric field at x = 0. magnitude V/m direction Determine the magnitude and direction of the electric field at x = 3.00 m. magnitude V/m direction

a)

When x =  0 m, V = 10.6 - 4.90*(0) = 10.6 V

When x =  10 m, V = 10.6 - 4.90*(10) = -38.4 V

When x =  6 m, V = 10.6 - 4.90*(6) = -18.8 V

When x =  3.00 m, V = 10.6 - 4.90*(3.00) = -4.1 V

b)

Explanation:

a) If V = a + bx , and a = 10.6 V and b = -4.90 V/m we have:

When x =  0 m, V = 10.6 - 4.90*(0) = 10.6 V

When x =  10 m, V = 10.6 - 4.90*(10) = -38.4 V

When x =  6 m, V = 10.6 - 4.90*(6) = -18.8 V

When x =  3.00 m, V = 10.6 - 4.90*(3.00) = -4.1 V

b) The electric field is:

In this particular case E depends only of b, it means E is a constant value. Therefore, E = 4.9 V/m when x = 0 m and when x = 3.00 m.

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards