# Video: Finding the Scalar Components at a Point of a Two-Dimensional Electric Field Vector

At a point π in space, the direction of the electric field vector is π = (1/β5)π’ β (2/β5)π£ and the magnitude of the electric field is 400.0 V/m. What is the scalar component π_π± of the field π at the point π? What is the scalar component π_π² of the field π at the point π? What is the scalar component π_π³ of the field π at the point π?

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### Video Transcript

At a point π in space, the direction of the electric field vector is π equals one over the square root of five π’ minus two over the square root of five π£. And the magnitude of the electric field is 400.0 volts per meter. What is the scalar component π π± of the field π at the point π? What is the scalar component π π² of the field π at the point π? What is the scalar component π π³ of the field π at the point π?

In this statement, weβre told the electric field direction as well as its magnitude. That is, weβre given all the information about this vector. We want to solve for its π±-, π²-, and π³-components at the point π in space. Letβs clear our workspace to begin doing that.

Knowing the direction that π takes at point π as well as its magnitude, we can begin solving for π sub π± by multiplying the magnitude and the direction together.

In particular, weβll use the π’-component of π which is its component in the π₯-direction. π sub π± equals that component multiplied by the overall magnitude of the vector π. When we multiply these terms together, we find that π sub π±, to four significant figures, is 178.9 volts per meter. Thatβs the π±-component of the electric field at point π.

To move on and solve for π sub π², this time instead of the π±-component of π, weβll focus on the π²- or π£-component. So in our equation, all weβll need to do the solve for π sub π² 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 π sub π² is negative 357.8 volts per meter. Thatβs the π²-component of the electric field at π. When we go to solve for the last component of π, π sub π³, when we look at our expression for the electric field, we see that thereβs no π€- or π³-component.

That means that π sub π³ is zero times the magnitude 400.0 volts per meter or simply zero point zero volts per meter. Thatβs the π³-component of the electric field π at point π.