Electrical power $P$ is given by

$P=\frac{E^2}{R},$

where $E$ is voltage and $R$ is resistance. Approximate the maximum percent error in calculating power when $120$ volts is applied to a $2,000$ ohm resistor and the possible percent errors in measuring $E$ and $R$ are $2\%$ and $3\%,$ respectively. $($Round your answer to the nearest percent.$)$

$\frac{dP}{P}=$

$\%$