Electric Field Physics Example 1

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Example 1

medium
What is the electric field 0.5 m0.5 \text{ m} from a point charge of 4×10−6 C4 \times 10^{-6} \text{ C}? Use k=9×109 N m2/C2k = 9 \times 10^9 \text{ N m}^2/\text{C}^2.

Solution

  1. 1
    Electric field from a point charge: E=kqr2E = k\frac{q}{r^2}
  2. 2
    E=9×109×4×10−6(0.5)2=9×109×4×10−60.25E = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.5)^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.25}
  3. 3
    E=9×109×1.6×10−5=1.44×105 N/CE = 9 \times 10^9 \times 1.6 \times 10^{-5} = 1.44 \times 10^5 \text{ N/C}

Answer

E=1.44×105 N/CE = 1.44 \times 10^5 \text{ N/C}
The electric field describes the force per unit charge at a point in space. It points radially outward from positive charges and inward toward negative charges.

About Electric Field

A region around a charged object where other charges experience a force. Measured in newtons per coulomb (N/C) or volts per meter (V/m).

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More Electric Field Examples

Example 2 medium

A charge of [formula] is placed in an electric field of [formula]. What force does it experience?

Example 3 hard

Two parallel plates are separated by [formula] with a potential difference of [formula]. What is the

Example 4 hard

Two point charges are placed on the x-axis: [formula] at [formula] and [formula] at [formula]. Find

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