At a point 𝑝 in space, the direction of the electric field vector is 𝐄 equals one over the square root of five 𝐢 minus two over the square root of five 𝐣. And the magnitude of the electric field is 400.0 volts per meter. What is the scalar component 𝐄 𝐱 of the field 𝐄 at the point 𝑝? What is the scalar component 𝐄 𝐲 of the field 𝐄 at the point 𝑝? What is the scalar component 𝐄 𝐳 of the field 𝐄 at the point 𝑝?
In this statement, we’re told the electric field direction as well as its magnitude. That is, we’re given all the information about this vector. We want to solve for its 𝐱-, 𝐲-, and 𝐳-components at the point 𝑝 in space. Let’s clear our workspace to begin doing that.
Knowing the direction that 𝐄 takes at point 𝑝 as well as its magnitude, we can begin solving for 𝐄 sub 𝐱 by multiplying the magnitude and the direction together.
In particular, we’ll use the 𝐢-component of 𝐄 which is its component in the 𝑥-direction. 𝐄 sub 𝐱 equals that component multiplied by the overall magnitude of the vector 𝐄. When we multiply these terms together, we find that 𝐄 sub 𝐱, to four significant figures, is 178.9 volts per meter. That’s the 𝐱-component of the electric field at point 𝑝.
To move on and solve for 𝐄 sub 𝐲, this time instead of the 𝐱-component of 𝐄, we’ll focus on the 𝐲- or 𝐣-component. So in our equation, all we’ll need to do the solve for 𝐄 sub 𝐲 is to replace one over the square root of five with negative two over the square root of five.
When we multiply these two numbers together, we find that 𝐄 sub 𝐲 is negative 357.8 volts per meter. That’s the 𝐲-component of the electric field at 𝑝. When we go to solve for the last component of 𝐄, 𝐄 sub 𝐳, when we look at our expression for the electric field, we see that there’s no 𝐤- or 𝐳-component.
That means that 𝐄 sub 𝐳 is zero times the magnitude 400.0 volts per meter or simply zero point zero volts per meter. That’s the 𝐳-component of the electric field 𝐄 at point 𝑝.