See main article: 1832 United States presidential election.
| Election Name: | 1832 United States presidential election in New Jersey |
| Country: | New Jersey |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1828 United States presidential election in New Jersey |
| Previous Year: | 1828 |
| Next Election: | 1836 United States presidential election in New Jersey |
| Next Year: | 1836 |
| Election Date: | November 2 – December 5, 1832 |
| Image1: | Andrew jackson head.jpg |
| Nominee1: | Andrew Jackson |
| Party1: | Democratic Party (United States) |
| Home State1: | Tennessee |
| Running Mate1: | Martin Van Buren |
| Electoral Vote1: | 8 |
| Popular Vote1: | 23,826 |
| Percentage1: | 49.89% |
| Nominee2: | Henry Clay |
| Party2: | National Republican Party (United States) |
| Home State2: | Kentucky |
| Running Mate2: | John Sergeant |
| Electoral Vote2: | 0 |
| Popular Vote2: | 23,466 |
| Percentage2: | 49.13% |
| Map Size: | 230px |
The 1832 United States presidential election in New Jersey took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
New Jersey voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won New Jersey by a margin of 0.76%. This was the closest election in the state's history.
| 1832 United States presidential election in New Jersey[1] | ||||||
|---|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | ||
| Democratic | Andrew Jackson (incumbent) | 23,826 | 49.89% | 8 | ||
| width: 3px" | National Republican | Henry Clay | 23,466 | 49.13% | 0 | |
| Anti-Masonic | William Wirt | 468 | 0.98% | 0 | ||
| Totals | 47,760 | 100.0% | 8 | |||