The electric potential at points in xy plane is given by V left parenthesis x comma y right parenthesis equals open parentheses 4 fraction numerator text v end text over denominator m squared end fraction close parentheses x squared minus open parentheses 9 fraction numerator text v end text over denominator m squared end fraction close parentheses y squared
What is the magnitude of the electric field at the point open parentheses 1 half semicolon 1 over 6 close parentheses
To find the magnitude of the electric field, we need to take the negative gradient of the electric potential function.
In this case, let's find the x-component and y-component of the electric field separately and then combine them to find the magnitude of the electric field.
The x-component of the electric field (Ex) is given by the partial derivative of the electric potential function with respect to x:
Ex = -∂V/∂x
To find ∂V/∂x, we differentiate the electric potential function with respect to x while treating y as a constant:
∂V/∂x = 8v/m² * x
Similarly, the y-component of the electric field (Ey) is given by the partial derivative of the electric potential function with respect to y:
Ey = -∂V/∂y
To find ∂V/∂y, we differentiate the electric potential function with respect to y while treating x as a constant:
∂V/∂y = -18v/m² * y
Now, let's calculate the x-component and y-component of the electric field at the point (1/2, 1/6):
Ex = 8v/m² * (1/2) = 4v/m²
Ey = -18v/m² * (1/6) = -3v/m²
To find the magnitude of the electric field (E), we use the Pythagorean theorem:
E = √(Ex² + Ey²) = √((4v/m²)² + (-3v/m²)²) = √(16v²/m^4 + 9v²/m^4) = √(25v²/m^4) = 5v/m²
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5v/m².
Calculate the magnitude of the electric field at a point P which is 30 cm to the right of a point charge Q equals negative 3 cross times 10 to the power of negative 6 end exponent space text C end text
To calculate the magnitude of the electric field at a point P, which is a distance of 30 cm to the right of a point charge Q, we can use Coulomb's Law.
Coulomb's Law states that the magnitude of the electric field E created by a point charge Q at a distance r from the charge is given by:
E = k * |Q| / r²
where k is the electrostatic constant (k ≈ 9 × 10^9 N m²/C²), |Q| is the magnitude of the charge, and r is the distance from the charge.
In this case, the charge Q is -3 × 10^(-6) C and the distance from the charge is 30 cm (which we will convert to meters).
Converting 30 cm to meters:
30 cm = 0.3 m
Now, we can substitute the values into the formula to calculate the magnitude of the electric field:
E = (9 × 10^9 N m²/C²) * |-3 × 10^(-6) C| / (0.3 m)²
E = (9 × 10^9 N m²/C²) * (3 × 10^(-6) C) / (0.3 m)²
E = (9 × 3) * (10^9 N m²/C² * 10^(-6) C) / (0.3 m)²
E = 27 * (10^3 N m²) / (0.09 m²)
E = 27 * 10^3 N / 0.09
E = 300,000 N/C
Therefore, the magnitude of the electric field at point P, which is 30 cm to the right of a point charge Q, is 300,000 N/C.
Consider a negative charge located between two parallel metal plates that sets up a uniform, downward electric field with magnitude E.
If the charge moves opposite the direction of the electric field,
If the charge moves opposite to the direction of the electric field, it means that it is moving in the opposite direction of the force exerted by the electric field.
In this case, the force on the negative charge will be in the opposite direction of the electric field. According to Newton's second law, the force on a particle is given by F = q * E, where F is the force, q is the charge, and E is the electric field.
Since the charge is negative, the force will be in the opposite direction of the electric field. Therefore, if the charge moves opposite to the direction of the electric field, it means that the force on the charge is acting in the same direction as its motion.
This indicates that there is an external force acting on the charge that is driving it against the electric field. This external force is responsible for overcoming the force exerted by the electric field and causing the charge to move in the opposite direction.
It is important to note that in this scenario, the charge is moving against the electric field due to the influence of an external force. In the absence of this external force, the charge would naturally move in the direction of the electric field.
To find the magnitude of the electric field at a point in the xy plane, we can take the negative gradient of the electric potential function V(x, y).
The electric field (E) is given by the equation: E = -∇V
Let's calculate the gradient of V(x, y):
∇V = ∂V/∂x * i + ∂V/∂y * j
Where i and j are the unit vectors in the x and y directions, respectively.
Taking the partial derivatives of V(x, y) with respect to x and y, we get:
∂V/∂x = 8v/m^2 * x
∂V/∂y = -18v/m^2 * y
Plugging these values into the gradient equation, we have:
∇V = 8v/m^2 * x * i - 18v/m^2 * y * j
To find the magnitude of the electric field at the point (1/2, 1/6), we substitute these coordinates into the equation:
E = sqrt[(8v/m^2 * (1/2))^2 + (-18v/m^2 * (1/6))^2]
Simplifying the expression inside the square root:
E = sqrt[(4v/m^2)^2 + (-3v/m^2)^2]
E = sqrt[16v^2/m^4 + 9v^2/m^4]
E = sqrt[(25v^2/m^4)]
E = (5v/m^2)
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5v/m^2.
To find the magnitude of the electric field at a given point, we need to take the negative gradient of the electric potential function.
The electric field E (magnitude) at a point (x, y) is given by:
E = - (∂V/∂x)î - (∂V/∂y)ĵ
Where î and ĵ are the unit vectors in the x and y directions respectively, and (∂V/∂x) and (∂V/∂y) are partial derivatives of the electric potential function with respect to x and y.
Let's find these partial derivatives and evaluate the electric field at the given point (1/2, 1/6):
Given V(x, y) = (4V/m^2)x^2 - (9V/m^2)y^2
Taking the partial derivative with respect to x, we get:
∂V/∂x = 8(V/m^2)x
Taking the partial derivative with respect to y, we get:
∂V/∂y = -18(V/m^2)y
Now, substitute the given point into these partial derivatives:
∂V/∂x = 8(V/m^2)(1/2) = 4(V/m^2)
∂V/∂y = -18(V/m^2)(1/6) = -3(V/m^2)
Now we can calculate the magnitude of the electric field at the given point:
E = - (∂V/∂x)î - (∂V/∂y)ĵ
= - (4(V/m^2))î - (-3(V/m^2))ĵ
= -4(V/m^2)î + 3(V/m^2)ĵ
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 4(V/m^2)î - 3(V/m^2)ĵ.