The electric potential in a certain region is V = a x2 + b x + c, where a = 15 V/m2, b = −13 V/m, and c = 67 V.
1)Determine the electric field at x = 4 m.
2)Determine the position where the electric field is zero.
Put given a,b,c,x values 1st problem is solved.2. To find 2nd put v= 0 and solve the quadratic equation.
1) Electric field can be calculated by taking the negative derivative of the electric potential with respect to x. So, let's find the derivative of V with respect to x first:
V = ax^2 + bx + c
dV/dx = 2ax + b
Now, let's substitute the given values for a and b:
dV/dx = 2(15 V/m^2)(4 m) + (-13 V/m)
= 120 V/m + (-13 V/m)
= 107 V/m
Therefore, the electric field at x = 4 m is 107 V/m in the positive direction.
2) To determine where the electric field is zero, we need to find the position(s) where dV/dx = 0. Let's set up the equation and solve it:
2ax + b = 0
Substituting the given values of a and b:
2(15 V/m^2)x + (-13 V/m) = 0
30x - 13 = 0
30x = 13
x = 13/30 m
So, the position where the electric field is zero is x = 13/30 m.
To determine the electric field at a given point or the position where the electric field is zero, we need to find the derivative of the electric potential function, V(x), with respect to x. The electric field, E, is equal to the negative derivative of the electric potential: E = -dV/dx.
1) To find the electric field at x = 4 m, we need to calculate -dV/dx at that point.
Given:
V = a x^2 + b x + c, where a = 15 V/m^2, b = -13 V/m, and c = 67 V.
First, let's find the derivative of V(x) with respect to x:
dV/dx = d/dx (a x^2 + b x + c)
= 2a x + b
Substituting the values of a and b:
dV/dx = 2(15) x - 13
= 30x - 13
Now, we can determine the electric field at x = 4 m by plugging in this value into the expression for the derivative:
E = -(30x - 13)
= -(30*4 - 13)
= -107 V/m
Therefore, the electric field at x = 4 m is -107 V/m.
2) To find the position where the electric field is zero, we need to determine the value of x at which the electric field E is equal to zero.
Setting the expression for the electric field E = 0, we have:
0 = 30x - 13
To solve for x, we isolate it by moving the constant term to the other side of the equation:
30x = 13
Dividing both sides by 30:
x = 13/30
Therefore, the position where the electric field is zero is x = 13/30 m.