See main article: 1804 United States presidential election.
| Election Name: | 1804 United States presidential election in New Jersey |
| Country: | New Jersey |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1800 United States presidential election in New Jersey |
| Previous Year: | 1800 |
| Next Election: | 1808 United States presidential election in New Jersey |
| Next Year: | 1808 |
| Election Date: | November 2 - December 5, 1804 |
| Image1: | Thomas Jefferson by Rembrandt Peale, 1800.jpg |
| Nominee1: | Thomas Jefferson |
| Party1: | Democratic-Republican Party |
| Home State1: | Virginia |
| Running Mate1: | George Clinton |
| Electoral Vote1: | 8 |
| Popular Vote1: | 13,119 |
| Percentage1: | 99.86% |
| President | |
| Before Election: | Thomas Jefferson |
| Before Party: | Democratic-Republican Party |
| After Election: | Thomas Jefferson |
| After Party: | Democratic-Republican Party |
The 1804 United States presidential election in New Jersey took place between November 2 and December 5, 1804, as part of the 1804 United States presidential election. The state chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, New Jersey cast eight electoral votes for Democratic Republican incumbent Thomas Jefferson.[1]