Two 1.60cm×1.60cm plates that form a parallel-plate capacitor are charged to ± 0.711nC . What is the potential difference across
the capacitor if the spacing between the plates is 1.20mm What is the potential difference across the capacitor if the spacing between the plates is 2.40mm ?
The surface charge density: S(igma) = Q / A A = 1.6 * 10^(-2) * 1.6 * 10^(-2) = 2.56 * 10^(-4) m² The electric field: E = S / E(psilon) o E = 0.711 * 10^^(-9) / ( 2.56 * 10 ^(-4) * 8.85 * 10 ^(-12) ) = = 0.3138 * 10 ^6 = 3.138 * 10^5 V/m a ) Δ V = E d Δ V = 3.138 * 10^5 V/m* 1.2 * 10^(-3) m = 376.56 V b ) When d = 2.40 mm: Δ V = 3.138 * 10^5 V/m * 2.40 *10^(-3) m = 753.12 V
The correct answer is: if is a root of , then is also a root of .
In fact, every polynomial has real and/or complex solutions. If all solutions are real, we're good. But if not all of them are real, then the complex ones come in couple of conjugate solutions. Since and are conjugate complex numbers, if one of them is a solution, the other must be as well.