Question

-A-Equilateral triangle, the length of the triangle edge is (1) these edges have a uniform linear charge density (A, A, 2A), Find the Electric field in the center of the

triangle. B- A parallel plate capacitor with a separation of 1 cm has 29 kV applied,

Answer 1

The electric field within an equilateral triangle with charged edges can be found using superposition, summing the vertical components due to charge symmetry. The electric field in a parallel plate capacitor is given by the potential difference divided by plate separation. The electric field would be 290,000 V/m.

Step-by-step explanation:

The student's question relates to the calculation of the electric field in various configurations, involving charged objects and electrical potentials. For the equilateral triangle with edges carrying different linear charge densities, we need to use superposition to calculate the resultant electric field at the center. Each side's charge density will produce an electric field directed towards or away from the center, depending on the charge's sign. By symmetry, the horizontal components will cancel out, and we will sum the vertical components to find the net field.

For the parallel plate capacitor question, the electric field (E) in a uniform field is given by E = V/d, where V is the potential difference and d is the separation between the plates. Hence, with a potential difference of 29 kV and a separation of 1 cm, the electric field would be 290,000 V/m.